I have heard from my team leader that this problem was supposed to be selected for BMO 2018 as problem 2. However, with one vote, A2 became selected for the contest.

Pretty much amusing cause I complicated myself a lot with this solution To solve this problem. I will use the following lemma Lemma 01. (Indonesian MO Mock #03 Problem) Given that $x,y,z$ are positive reals such that $xyz = 1$. Therefore, \[ (x + y + z)^2 \left( \frac{1}{x^2} + \frac{1}{y^2} + \frac{1}{z^2} \right) \ge 9 + 2 ( x^3 + y^3 + z^3 ) + 4 \left( \frac{1}{x^3} + \frac{1}{y^3} + \frac{1}{z^3 } \right) \]Proof. Expand $\left( (x - y)(y - z)(z - x) \right)^2 \ge 0 $ and laugh Now, notice that by our lemma, \[ 2(a^2 + b^2 + c^2) \left( \frac{1}{a^2} + \frac{1}{b^2} + \frac{1}{c^2} \right) \ge \frac{2}{3} (a + b + c)^2 \left( \frac{1}{a^2} + \frac{1}{b^2} + \frac{1}{c^2} \right) \ge 6 + \frac{4}{3} (a^3 + b^3 + c^3) + \frac{8}{3} \left( \frac{1}{a^3} + \frac{1}{b^3} + \frac{1}{c^3} \right) \] It suffices to prove that \[ 18 + 4( a^3 + b^3 + c^3) + 8 \left( \frac{1}{a^3} + \frac{1}{b^3} + \frac{1}{c^3} \right) \ge 9 \left( a + b + c + \frac{1}{a} + \frac{1}{b} + \frac{1}{c} \right) \] This is pretty much obvious by the fact that by AM GM, \[ 18 + 4 \sum_{cyc} a^3 + 8 \sum_{cyc} \frac{1}{a^3} \ge 3 \sum_{cyc} \left( \frac{1}{a^3} + 2 \right) + 3 \sum_{cyc} ( a^3 + 2 ) \ge 9 \sum_{cyc} \left( a + \frac{1}{a}\right) \] Remark. I honestly agree that my solution is really complicated. (There's a much simpler solution like @below. But as i saw this problem, I recall the lemma and used it to help my bash (for not wasting my time for a better solution), which is then a lot faster.

sqing wrote: Let $ a, b, c$ be positive real numbers such that $ abc = 1. $ Prove that: $$ 2 (a^ 2 + b^ 2 + c^ 2) \left (\frac 1 {a^ 2} + \frac 1{b^ 2}+ \frac 1{c^2}\right)\geq 3(a+ b + c + ab + bc + ca).$$ $$(a^2 + b ^2 + c ^2) \left (\frac 1 {a^ 2} + \frac 1{b^ 2}+ \frac 1{c^2}\right)\geq 3(a^2 + b ^2 + c ^2) \geq 3 (a + b + c)$$$$\left (\frac 1 {a^ 2} + \frac 1{b^ 2}+ \frac 1{c^2}\right)(a^2 + b ^2 + c ^2) \geq 3\left (\frac 1 {a^ 2} + \frac 1{b^ 2}+ \frac 1{c^2}\right) \geq 3 \left (\frac 1 {a} + \frac 1{b}+ \frac 1{c}\right)=3( ab + bc + ca)$$Adding the last two inequalities we get the required.

This inequality was proposed by me for the BMO short list 2018.

florin_rotaru wrote: This inequality was proposed by me for the BMO short list 2018. Congratulations. Given $a,b,c>0, abc=1.$ Prove that$$(a+ \frac{1}{b})^2+(b+ \frac{1}{c})^2+(c+ \frac{1}{a})^2 \geq 3(a+b+c+1).$$

Thank you.

Problem BMO 2018 Let $ a $,$ b$,$ c$, be positive real numbers such that $abc= 1 $. Show that $$2(a^{2}+b^{2}+c^{2})(\frac{1}{a^{2}}+\frac{1}{b^{2}}+\frac{1}{c^{2}})\geq 3(a+b+c+ab+bc+ca) . $$ Solution We will first show that $$2(\frac{a}{b}+\frac{b}{c}+\frac{c}{a})\geq a+\frac{1}{a}+b+\frac{1}{b}+c+\frac{1}{c}, (1) $$ We will show that $$ \frac{a}{b}+\frac{b}{c}+\frac{c}{a}\geq \frac{1}{a}+\frac{1}{b}+\frac{1}{c} \ and \ \frac{a}{b}+\frac{b}{c}+\frac{c}{a}\geq a+b+c $$ which sum up leads to inequality (1). We have inequality $$ \frac{a}{b}+\frac{a}{b}+\frac{c}{a}\geq 3\frac{\sqrt[3]{abc}}{b}$$ Analogous : $$\frac{b}{c}+\frac{b}{c}+\frac{a}{b}\geq 3\frac{\sqrt[3]{abc}}{c} $$and $$ \frac{c}{a}+\frac{c}{a}+\frac{b}{c}\geq 3\frac{\sqrt[3]{abc}}{a} $$ Summing these inequalities results $$ 3(\frac{a}{b}+\frac{b}{c}+\frac{c}{a})\geq 3\sqrt[3]{abc}(\frac{1}{a}+\frac{1}{b}+\frac{1}{c})\Leftrightarrow \frac{a}{b}+\frac{b}{c}+\frac{c}{a}\\\geq \sqrt[3]{abc}(\frac{1}{a}+\frac{1}{b}+\frac{1}{c})\geq \frac{1}{a}+\frac{1}{b}+\frac{1}{c} $$well $$ \frac{a}{b}+\frac{b}{c}+\frac{c}{a}\geq \frac{1}{a}+\frac{1}{b}+\frac{1}{c},(2) $$ We have inequality $$ \frac{a}{b}+\frac{a}{b}+\frac{b}{c}\geq 3\frac{\sqrt[3]{a^{2}}}{\sqrt[3]{bc}}=\frac{3a}{\sqrt[3]{abc}} $$ Analogous : $$ \frac{b}{c}+\frac{b}{c}+\frac{c}{a}\geq \frac{3b}{\sqrt[3]{abc}}$$and $$ \frac{c}{a}+\frac{c}{a}+\frac{a}{b}\geq \frac{3c}{\sqrt[3]{abc}} $$. Summing these inequalities results $$ 3(\frac{a}{b}+\frac{b}{c}+\frac{c}{a})\geq \frac{3(a+b+c)}{\sqrt[3]{abc}}\Leftrightarrow \frac{a}{b}+\frac{b}{c}+\frac{c}{a}\geq \frac{a+b+c}{\sqrt[3]{abc}}=a+b+c $$well $$ \frac{a}{b}+\frac{b}{c}+\frac{c}{a}\geq a+b+c,(3) $$ Summing relationships (2) and (3) inequality results (1) . Result $$ 6(\frac{a}{b}+\frac{b}{c}+\frac{c}{a})\geq 3(a+\frac{1}{a}+b+\frac{1}{b}+c+\frac{1}{c}), (4) $$ With the inequality of Cauchy-Buniakovski-Schwarz we have : $$ (a^{2}+b^{2}+c^{2})(\frac{1}{a^{2}}+\frac{1}{b^{2}}+\frac{1}{c^{2}})=(a^{2}+b^{2}+c^{2})(\frac{1}{b^{2}}+\frac{1}{c^{2}}+\frac{1}{a^{2}})\geq (\frac{a}{b}+\frac{b}{c}+\frac{c}{a})^{2} $$ So we have : $$ 2(a^{2}+b^{2}+c^{2})(\frac{1}{a^{2}}+\frac{1}{b^{2}}+\frac{1}{c^{2}})=(a^{2}+b^{2}+c^{2})(\frac{1}{a^{2}}+\frac{1}{b^{2}}+\frac{1}{c^{2}})+(a^{2}+b^{2}+c^{2})(\frac{1}{a^{2}}+\frac{1}{b^{2}}+\frac{1}{c^{2}})\geq (\frac{a}{b}+\frac{b}{c}+\frac{c}{a})^{2}+9 $$well $$ 2(a^{2}+b^{2}+c^{2})(\frac{1}{a^{2}}+\frac{1}{b^{2}}+\frac{1}{c^{2}})\geq (\frac{a}{b}+\frac{b}{c}+\frac{c}{a})^{2}+9 $$ Summing up (4) and (5) results $$ 6(\frac{a}{b}+\frac{b}{c}+\frac{c}{a})+2(a^{2}+b^{2}+c^{2})(\frac{1}{a^{2}}+\frac{1}{b^{2}}+\frac{1}{c^{2}})\geq 3(a+\frac{1}{a}+b+\frac{1}{b}+c+\frac{1}{c})+(\frac{a}{b}+\frac{b}{c}+\frac{c}{a})^{2}+9 \Leftrightarrow $$$$ 2(a^{2}+b^{2}+c^{2})(\frac{1}{a^{2}}+\frac{1}{b^{2}}+\frac{1}{c^{2}})-3(a+\frac{1}{a}+b+\frac{1}{b}+c+\frac{1}{c})\geq (\frac{a}{b}+\frac{b}{c}+\frac{c}{a})^{2}-6(\frac{a}{b}+\frac{b}{c}+\frac{c}{a})+9=(\frac{a}{b}+\frac{b}{c}+\frac{c}{a}-3)^{2}\geq 0 $$ Resulting inequality $$ 2(a^{2}+b^{2}+c^{2})(\frac{1}{a^{2}}+\frac{1}{b^{2}}+\frac{1}{c^{2}})\geq 3(a+\frac{1}{a}+b+\frac{1}{b}+c+\frac{1}{c})=3(a+b+c+ab+bc+ca) $$i.e inequality $$ 2(a^{2}+b^{2}+c^{2})(\frac{1}{a^{2}}+\frac{1}{b^{2}}+\frac{1}{c^{2}})\geq3(a+b+c+ab+bc+ca) $$ Equality occurs if and only if $$ a = b = c = 1 . $$

A rather simple solution. As $a+b+c \geq3$ and by Tito Andrescu we have $a^2+b^2+c^2\geq (a+b+c)^2/3\geq a+b+c (1)$ as $$1/a^2=b^2c^2$$$$1/b^2=a^2c^2$$$$1/c^2=b^2a^2$$As $ab+bc+ac\geq 3$ And then applying $(1)$ We can say $2(a^2+b^2+c^2)*(1/a^2+1/b^2+1/c^2)\geq 2(a+b+c)(ab+bc+ac)$ Let $a+b+c=3+k$ and $ab+bc+ac=3+m$ It is enough to prove $2(3+k)(3+m)\geq3*(6+k+m)$ $18+6k+6m+km\geq 18+3k+3m$ Which is obviously true as $k,m$ are non negative. Equality holds for $k=0 m=0$ therefore $a=b=c=1$ Or did I make a mistake somewhere?

sqing wrote: Let $ a, b, c$ be positive real numbers such that $ abc = 1. $ Prove that: $$ 2 (a^ 2 + b^ 2 + c^ 2) \left (\frac 1 {a^ 2} + \frac 1{b^ 2}+ \frac 1{c^2}\right)\geq 3(a+ b + c + ab + bc + ca).$$ Solution 1: $$LHS \geq 6(a^2+b^2+c^2)\sqrt[3]{\frac{1}{a^2b^2c^2}}=6(a^2+b^2+c^2)\geq 3\left(\frac{(a+b+c)^2}{3}+ab+bc+ca \right)\geq 3(a+b+c+ab+bc+ca)$$

sqing wrote: Let $ a, b, c$ be positive real numbers such that $ abc = 1. $ Prove that: $$ 2 (a^ 2 + b^ 2 + c^ 2) \left (\frac 1 {a^ 2} + \frac 1{b^ 2}+ \frac 1{c^2}\right)\geq 3(a+ b + c + ab + bc + ca).$$ Solution 2: Using pqr notations: $p =a+b+c\geq 3$ and $q =ab+bc+ca\geq 3$. Therefore: $$LHS \geq \frac{2p^2q^2}{9} \geq 2pq \geq 3(p+q) = RHS$$

I hope this is correct. First, considering that $abc=1$ , we put $a=x/y, b=y/z, c=z/y$. Then after expanding we have; $2\frac{(x^4z^2+x^2y^4+z^4y^2)(y^4z^2+x^2z^4+x^4y^2)}{(xyz)^3} \geq \sum_{sym}x^2y$ After multiplying by $(xyz)^3$ and expanding again we have: $6x^4y^4z^4+2(x^6y^6+y^6z^6+x^6y^6)+2(x^8y^2z^2+x^2y^8z^2+x^2y^2z^8) \geq \sum_{sym}x^5y^4z^3$} Now, by Murihead we have : $\sum_{sym}x^6z^6 \geq \sum_{sym}x^5y^4z^3$ because $(6,6,0) \succ (5,4,3)$,. All terms on the LHS are positive, so with the inequality presented above we are done.

Note that $a^2+b^2+c^2\ge a+b+c$ is true since $(2,0,0)\succ\left(\frac43,\frac13,\frac13\right)$. Also, $a^2+b^2+c^2\ge3(abc)^{2/3}=3$ and $\frac1{a^2}+\frac1{b^2}+\frac1{c^2}\ge3(abc)^{-2/3}=3$. Then we finish by adding the following inequalities: $$(a^2+b^2+c^2)\left(\frac1{a^2}+\frac1{b^2}+\frac1{c^2}\right)\ge3(a^2+b^2+c^2)\ge3(a+b+c)$$and $$(a^2+b^2+c^2)\left(\frac1{a^2}+\frac1{b^2}+\frac1{c^2}\right)\ge3\left(\frac1{a^2}+\frac1{b^2}+\frac1{c^2}\right)\ge3\left(\frac1a+\frac1b+\frac1c\right)=3(ab+bc+ca).$$

programming_wiz wrote: I hope this is correct. First, considering that $abc=1$ , we put $a=x/y, b=y/z, c=z/y$. Then after expanding we have; $2\frac{(x^4z^2+x^2y^4+z^4y^2)(y^4z^2+x^2z^4+x^4y^2)}{(xyz)^3} \geq \sum_{sym}x^2y$ After multiplying by $(xyz)^3$ and expanding again we have: $6x^4y^4z^4+2(x^6y^6+y^6z^6+x^6y^6)+2(x^8y^2z^2+x^2y^8z^2+x^2y^2z^8) \geq \sum_{sym}x^5y^4z^3$} Now, by Murihead we have : $\sum_{sym}x^6z^6 \geq \sum_{sym}x^5y^4z^3$ because $(6,6,0) \succ (5,4,3)$,. All terms on the LHS are positive, so with the inequality presented above we are done. Incorrect solution, but can be finished. Here's my solution which is very similar to your approach: Using the Ravi substitution because $abc=1$ we can substitute $a=\frac{x}{y}$, $b=\frac{y}{z}$, $c=\frac{z}{x}$. We'll use the notation: $$T[a_{1},a_{2},a_{3}]=\sum\limits_{sym} x^{a_{1}}y^{a_{2}}z^{a_{3}}$$throughout the solution (note that for example $T[1,0,0]=2x+2y+2z$ and $T[1,1,1]=6xyz$). Then:

(for $r=1$) in the symmetrical sum form we have that $T[3,0,0]+T[1,1,1]\geq 2T[2,1,0]$, so multiplying the inequality by $x^3y^3z^3$ we have that $T[6,3,3]+T[4,4,4]\geq 2T[5,4,3]$. Now, since $(8,2,2)\succ (5,4,3)$ and $(6,6,0)\succ (6,3,3)\succ (5,4,3)$ we have that: $$T[4,4,4]+T[6,6,0]+T[8,2,2]\geq T[4,4,4]+T[6,3,3]+T[8,2,2]\geq 2T[5,4,3]+T[8,2,2]\geq 3T[5,4,3]$$so we've proven the inequality.

I think easy for A4 $2(a^2 + b^2 + c^2) \left (\frac{1}{a^2} + \frac{1}{b^2} + \frac{1}{c^2} \right)=2(a^2 + b^2 + c^2) \left(b^2c^2+a^2c^2+a^2b^2 \right) \ge 2(a^2+b^2+c^2)(a+b+c) \ge 2(ab+bc+ca)(a+b+c)$ Let $ab+bc+ca=x , a+b+c=y$ Note that $x,y \ge 3(*)$ since $abc=1$ Thus we just need to show $2xy \ge 3x+3y \iff 2 \ge \frac{3}{x} + \frac{3}{y}$ which is obvious by $(*)$ ... so we are done

Hello First part $\frac1{a^2}+\frac1{b^2}+\frac1{c^2} \geq 3$ $6(a^2+b^2+c^2)\geq 3(a+b+c+ab+bc+ca)$ $2(a^2+b^2+c^2)\geq (a+b+c+ab+bc+ca)$ $a^2+b^2+c^2 \geq ab+bc+ca$ is well known lemma $a^2+b^2+c^2 \geq (a+b+c)(a+b+c)/3 \geq a+b+c$ And we are done:))

Ibrahim_K wrote: I think easy for A4 $2(a^2 + b^2 + c^2) \left (\frac{1}{a^2} + \frac{1}{b^2} + \frac{1}{c^2} \right)=2(a^2 + b^2 + c^2) \left(b^2c^2+a^2c^2+a^2b^2 \right) \ge 2(a^2+b^2+c^2)(a+b+c) \ge 2(ab+bc+ca)(a+b+c)$ Let $ab+bc+ca=x , a+b+c=y$ Note that $x,y \ge 3(*)$ since $abc=1$ Thus we just need to show $2xy \ge 3x+3y \iff 2 \ge \frac{3}{x} + \frac{3}{y}$ which is obvious by $(*)$ ... so we are done Good one ))

Let $ a, b, c$ be positive real numbers such that $ abc = 1. $ Prove that $$ 2 (a + b + c) \left (\frac 1 {a^ 2} + \frac 1{b^ 2}+ \frac 1{c^2}\right)\geq 3(a+ b + c + ab + bc + ca)$$$$ 2 (a^ 2 + b^ 2 + c^ 2) \left (\frac 1 {a} + \frac 1{b}+ \frac 1{c}\right)\geq 3(a+ b + c + ab + bc + ca) $$

Second one $6(a^2+b^2+c^2)\geq3(a+b+c+ab+bc+ac)$ Easy to prove

The first one we can say a+b+c=3 And We have to prove $a+b+c\geq ab+bc+ca$ By cauchy $(a+b+c)^2\geq 3(ab+bc+ca)$ which is true

sqing wrote: Let $ a, b, c$ be positive real numbers such that $ abc = 1. $ Prove that $$ 2 (a + b + c) \left (\frac 1 {a^ 2} + \frac 1{b^ 2}+ \frac 1{c^2}\right)\geq 3(a+ b + c + ab + bc + ca)$$$$ 2 (a^ 2 + b^ 2 + c^ 2) \left (\frac 1 {a} + \frac 1{b}+ \frac 1{c}\right)\geq 3(a+ b + c + ab + bc + ca) $$

sqing wrote:

Not at all.

Marinchoo wrote: programming_wiz wrote: I hope this is correct. First, considering that $abc=1$ , we put $a=x/y, b=y/z, c=z/y$. Then after expanding we have; $2\frac{(x^4z^2+x^2y^4+z^4y^2)(y^4z^2+x^2z^4+x^4y^2)}{(xyz)^3} \geq \sum_{sym}x^2y$ After multiplying by $(xyz)^3$ and expanding again we have: $6x^4y^4z^4+2(x^6y^6+y^6z^6+x^6y^6)+2(x^8y^2z^2+x^2y^8z^2+x^2y^2z^8) \geq \sum_{sym}x^5y^4z^3$} Now, by Murihead we have : $\sum_{sym}x^6z^6 \geq \sum_{sym}x^5y^4z^3$ because $(6,6,0) \succ (5,4,3)$,. All terms on the LHS are positive, so with the inequality presented above we are done. Incorrect solution, but can be finished. Here's my solution which is very similar to your approach: Using the Ravi substitution because $abc=1$ we can substitute $a=\frac{x}{y}$, $b=\frac{y}{z}$, $c=\frac{z}{x}$. We'll use the notation: $$T[a_{1},a_{2},a_{3}]=\sum\limits_{sym} x^{a_{1}}y^{a_{2}}z^{a_{3}}$$throughout the solution (note that for example $T[1,0,0]=2x+2y+2z$ and $T[1,1,1]=6xyz$). Then:

(for $r=1$) in the symmetrical sum form we have that $T[3,0,0]+T[1,1,1]\geq 2T[2,1,0]$, so multiplying the inequality by $x^3y^3z^3$ we have that $T[6,3,3]+T[4,4,4]\geq 2T[5,4,3]$. Now, since $(8,2,2)\succ (5,4,3)$ and $(6,6,0)\succ (6,3,3)\succ (5,4,3)$ we have that: $$T[4,4,4]+T[6,6,0]+T[8,2,2]\geq T[4,4,4]+T[6,3,3]+T[8,2,2]\geq 2T[5,4,3]+T[8,2,2]\geq 3T[5,4,3]$$so we've proven the inequality. LOVED YOUR SOLUTION

Note that all of these inequalities are essentially the same and I'm solving all of them because I'm bored sqing wrote: Let $ a, b, c$ be positive real numbers such that $ abc = 1. $ Prove that: $$ 2 (a^ 2 + b^ 2 + c^ 2) \left (\frac 1 {a^ 2} + \frac 1{b^ 2}+ \frac 1{c^2}\right)\geq 3(a+ b + c + ab + bc + ca).$$ Let $a=x^3, b=y^3, c=z^3\implies xyz=1$ $2\cdot LHS=4(a^2+b^2+c^2)\left(\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}\right)=4(x^6+y^6+z^6)\left(\frac{1}{x^6}+\frac{1}{y^6}+\frac{1}{z^6}\right)=4(x^6+y^6+z^6)(x^6y^6+y^6z^6+z^6x^6)=4S(12,6,0)+2S(6,6,6)$ $2\cdot RHS=6(a+b+c+ab+bc+ca)=6(x^3+y^3+z^3+x^3y^3+y^3z^3+z^3x^3)=6(x^8y^5z^5+x^5y^8z^5+x^5y^5z^8+x^7y^7z^4+x^7y^4z^7+x^4y^7z^7)=3S(8,5,5)+3S(7,7,4)$ Thus it is sufficient to prove $4S(12,6,0)+2S(6,6,6)\geq3S(8,5,5)+3S(7,7,4)$ From AM-GM and Murihead $4S(12,6,0)+2S(6,6,6)=3S(12,6,0)+(S(12,6,0)+S(6,6,6)+S(6,6,6))\geq3S(12,6,0)+3S(8,6,4)\geq3S(8,5,5)+3S(7,7,4)$ sqing wrote: Let $ a, b, c$ be positive real numbers such that $ abc = 1. $ Prove that $$ 2 (a + b + c) \left (\frac 1 {a^ 2} + \frac 1{b^ 2}+ \frac 1{c^2}\right)\geq 3(a+ b + c + ab + bc + ca)$$$$ 2 (a^ 2 + b^ 2 + c^ 2) \left (\frac 1 {a} + \frac 1{b}+ \frac 1{c}\right)\geq 3(a+ b + c + ab + bc + ca) $$ Let $a=x^3, b=y^3, c=z^3\implies xyz=1$ $2\cdot LHS=4(a+b+c)\left(\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}\right)=4(x^3+y^3+z^3)\left(\frac{1}{x^6}+\frac{1}{y^6}+\frac{1}{z^6}\right)=4(x^3+y^3+z^3)(x^6y^6+y^6z^6+z^6x^6)=4S(9,6,0)+2S(6,6,3)$ $2\cdot RHS=6(a+b+c+ab+bc+ca)=6(x^3+y^3+z^3+x^3y^3+y^3z^3+z^3x^3)=6(x^7y^4z^4+x^4y^7z^4+x^4y^4z^7+x^6y^6z^3+x^6y^3z^6+x^3y^6z^6)=3S(7,4,4)+3S(6,6,3)$ Thus it is sufficient to prove $4S(9,6,0)+2S(6,6,3)\geq3S(7,4,4)+3(6,6,3)\iff 4S(9,6,0)\geq3S(7,4,4)+S(6,6,3)$ which is true from Murihead Let $a=x^3, b=y^3, c=z^3\implies xyz=1$ $2\cdot LHS=4(a^2+b^2+c^2)\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)=4(x^6+y^6+z^6)\left(\frac{1}{x^3}+\frac{1}{y^3}+\frac{1}{z^3}\right)=4(x^6+y^6+z^6)(x^3y^3+y^3z^3+z^3x^3)=4S(9,3,0)+2S(6,3,3)$ $2\cdot RHS=6(a+b+c+ab+bc+ca)=6(x^3+y^3+z^3+x^3y^3+y^3z^3+z^3x^3)=6(x^6y^3z^3+x^3y^6z^3+x^3y^3z^6+x^5y^5z^2+x^5y^2z^5+x^2y^5z^5)=3S(6,3,3)+3S(5,5,2)$ Thus it is sufficient to prove $4S(9,3,0)+2S(6,3,3)\geq3S(6,3,3)+3(5,5,2)\iff 4S(9,3,0)\geq3S(5,5,2)+S(6,3,3)$ which is true from Murihead Hello_Ok wrote: (Stronger) Let $ a, b, c$ be positive real numbers such that $ abc = 1. $ Prove that $$ 2 (a^2 + b^2 + c^2) \left (\frac 1 {a^ 2} + \frac 1{b^ 2}+ \frac 1{c^2}\right)\geq 3(a+ b + c + a^2 + b^2 + c^2)$$$$ 2 (a^ 2 + b^ 2 + c^ 2) \left (\frac 1 {a^2} + \frac 1{b^2}+ \frac 1{c^2}\right)\geq 3(a^2+ b^2 + c^2 + ab + bc + ca) $$Note that these two inequalities are very easy , so what're your simple smart solutions? Let $a=x^3, b=y^3, c=z^3\implies xyz=1$ $2\cdot LHS=4(a^2+b^2+c^2)\left(\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}\right)=4(x^6+y^6+z^6)\left(\frac{1}{x^6}+\frac{1}{y^6}+\frac{1}{z^6}\right)=4(x^6+y^6+z^6)(x^6y^6+y^6z^6+z^6x^6)=4S(12,6,0)+2S(6,6,6)$ $2\cdot RHS=6(a+b+c+a^2+b^2+c^2)=6(x^3+y^3+z^3+x^6+y^6+z^6)=6(x^8y^5z^5+x^5y^8z^5+x^5y^5z^8+x^4y^4z^10+x^4y^10z^4+x^10y^4z^4)=3S(8,5,5)+3S(10,4,4)$ Thus it is sufficient to prove $4S(12,6,0)+2S(6,6,6)\geq3S(8,5,5)+3(10,4,4)$ From AM-GM and Murihead $4S(12,6,0)+2S(6,6,6)=3S(12,6,0)+(S(12,6,0)+S(6,6,6)+S(6,6,6))\geq3S(12,6,0)+3S(8,6,4)\geq3S(8,5,5)+3S(10,4,4)$ Let $a=x^3, b=y^3, c=z^3\implies xyz=1$ $2\cdot LHS=4(a^2+b^2+c^2)\left(\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}\right)=4(x^6+y^6+z^6)\left(\frac{1}{x^6}+\frac{1}{y^6}+\frac{1}{z^6}\right)=4(x^6+y^6+z^6)(x^6y^6+y^6z^6+z^6x^6)=4S(12,6,0)+2S(6,6,6)$ $2\cdot RHS=6(ab+bc+ca+a^2+b^2+c^2)=6(x^3y^3+y^3z^3+z^3x^3+x^6+y^6+z^6)=6(x^4y^7z^7+x^4y^7z^7+x^7y^7z^4+x^4y^4z^10+x^4y^10z^4+x^10y^4z^4)=3S(7,7,4)+3S(10,4,4)$ Thus it is sufficient to prove $4S(12,6,0)+2S(6,6,6)\geq3S(7,7,4)+3(10,4,4)$ From AM-GM and Murihead $4S(12,6,0)+2S(6,6,6)=3S(12,6,0)+(S(12,6,0)+S(6,6,6)+S(6,6,6))\geq3S(12,6,0)+3S(8,6,4)\geq3S(7,7,4)+3S(10,4,4)$ Hello_Ok wrote: (And stronger) Let $ a, b, c$ be positive real numbers such that $ abc = 1. $ Prove that $$ 2 (a + b+ c) \left (\frac 1 {a^ 2} + \frac 1{b^ 2}+ \frac 1{c^2}\right)\geq 3(a+ b + c + a^2 + b^2 + c^2)$$$$ 2 (a + b + c) \left (\frac 1 {a^2} + \frac 1{b^2}+ \frac 1{c^2}\right)\geq 3(a^2+ b^2 + c^2 + ab + bc + ca) $$Note that these two inequalities are very easy , so what're your simple smart solutions? Let $a=x^3, b=y^3, c=z^3\implies xyz=1$ $2\cdot LHS=4(a+b+c)\left(\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}\right)=4(x^3+y^3+z^3)\left(\frac{1}{x^6}+\frac{1}{y^6}+\frac{1}{z^6}\right)=4(x^3+y^3+z^3)(x^6y^6+y^6z^6+z^6x^6)=4S(9,6,0)+2S(6,6,3)$ $2\cdot RHS=6(a+b+c+a^2+b^2+c^2)=6(x^3+y^3+z^3+x^6+y^6+z^6)=6(x^7y^4z^4+x^4y^7z^4+x^4y^4z^7+x^3y^3z^9+x^3y^9z^3+x^9y^3z^3)=3S(7,4,4)+3S(9,3,3)$ Thus it is sufficient to prove $4S(9,6,0)+2S(6,6,3)\geq3S(7,4,4)+3(9,3,3)$ which is true from Murihead Let $a=x^3, b=y^3, c=z^3\implies xyz=1$ $2\cdot LHS=4(a+b+c)\left(\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}\right)=4(x^3+y^3+z^3)\left(\frac{1}{x^6}+\frac{1}{y^6}+\frac{1}{z^6}\right)=4(x^3+y^3+z^3)(x^6y^6+y^6z^6+z^6x^6)=4S(9,6,0)+2S(6,6,3)$ $2\cdot RHS=6(ab+bc+ca+a^2+b^2+c^2)=6(x^3y^3+y^3z^3+z^3x^3+x^6+y^6+z^6)=6(x^6y^6z^3+x^6y^3z^6+x^3y^6z^6+x^3y^3z^9+x^3y^9z^3+x^9y^3z^3)=3S(6,6,3)+3S(9,3,3)$ Thus it is sufficient to prove $4S(9,6,0)+2S(6,6,3)\geq3S(6,6,3)+3(9,3,3)\iff4S(9,6,0)\geq3S(9,3,3)+S(6,6,3)$ which is true from Murihead

Let $a,b,c>0$ and $abc=1$ .Prove that $$(a^2+b^2+c^2)\left(\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}- \frac{57}{10}\right)\ge -\frac{81}{10} $$

Generalization 1 Let $ a_{1}, a_{2}, a_{3},n$ ($n\geq 2$) be positive reels such that $a_{1}a_{2}a_{3}=1$. Then prove that $$ n\left(a_{1}^{n} +a_{2}^{n} + a_{3}^{n}\right) \left (\dfrac 1 {a_{1}^{n}} + \dfrac 1{a_{2}^{n}}+ \dfrac 1{a_{3}^{n}}\right)\geq 3\left(a_{1}^{n-1}+ a_{2}^{n-1} + a_{3}^{n-1} + \sum_{p=1,i=n-1}^{p=3,i=1}{\left(a_{p}^{i}a_{p+1}^{n-i}\right)}\right)$$

Generalization 2 Let $ a_{1},a_{2},\cdots,n,k$ be positive reels ($n,k\geq 2$) such that $\prod{a_{1}}=1$. Then prove that $$ n\left(\sum_{cyc}{a_{1}^{n}}\right) \left (\sum_{cyc}{\dfrac 1 {a_{1}^{n}}}\right)\geq k\left(\sum_{cyc}{a_{1}^{n-1}}+ \sum_{p=1,i=n-1}^{p=k,i=1}{\left(a_{p}^{i}a_{p+1}^{n-i}\right)}\right)$$

First open the brackets. 2*(3+a²/b²+a²/c²+b²/a²+c²/a²+c²/b²+b²/c²)>=3(ab+ac+bc+a+b+c) Then open the brackets again and apply AM-GM for two of them which of them are equal; a²/b²+a²/b²+b²+b²>= 4a a²/c+a²/c+c²+c²>=4a >=4b >=4b >=4c >=4c Then we will get; 6+2*(a²/b²+b²/c²+b²/a²+c²/a²+c²/b²+a²/c²)>=8a+8b+8c>=3a+3b+3c+3ab+3ac+3bc We proved the LHS. And if L>=H, H>=R then L>=R. From there , we must prove; 8a+8b+8c>=3a+3b+3c+3ab+3ac+3bc. Make it more simple; 5a+5b+5c>=3ab+3ac+3bc If abc=1 so, (a+b+c)/3>=root of abc (3 degree)=1 a+b+c>=3 5(a+b+c)>=15 15>=3(ab+ac+bc) 5>=ab+ac+bc Muirhead (2,0,0)>=(1,1,0) 2(a²+b²+c²)>=2*(ab+ac+bc) a²+b²+c²>=ab+ac+bc root of( (a²+b²+c²)/3 )>=root of( abc) (3 degree)=1 a²+b²+c²>=3 If ab+ac+bc is smaller than 5, then it must be smaller than 3 too. So we proved.

Systematicworker wrote: I have heard from my team leader that this problem was supposed to be selected for BMO 2018 as problem 2. However, with one vote, A2 became selected for the contest. It is interesting whether some Balkan/Junior Balkan actual paper will have a Schurhead-solve symmetric inequality and people who do not know this sym inequality method would be screwed So far I have seen a few unsuccessful attempts on voting such a problem, but who knows, it might actually become a reality in the near future.

Here is a nice solution by @africanboy during JBMO training. We may set $a = \frac{x}{y}$, $b = \frac{y}{z}$, $c = \frac{z}{x}$. The inequality is equivalent to \[ 6 + 2\left(\frac{x^2y^2}{z^4} + \frac{y^2z^2}{x^4} + \frac{x^2z^2}{y^4} + \frac{x^4}{y^2z^2} + \frac{y^4}{x^2z^2} + \frac{z^4}{x^2y^2}\right) \geq 3\left( \frac{x}{y} + \frac{y}{z} + \frac{z}{x} + \frac{y}{x} + \frac{z}{y} + \frac{x}{z}\right).\] By AM-GM we have $1 + \frac{x^4}{y^2z^2} + \frac{x^2z^2}{y^4} \geq 3\frac{x^2}{z^2}$. Together with the other five analogous inequalities, we obtain that the left-hand side is at least $3\left(\frac{x^2}{y^2} + \frac{y^2}{z^2} + \frac{z^2}{x^2} + \frac{y^2}{x^2} + \frac{z^2}{y^2} + \frac{x^2}{z^2}\right)$. Now it just remains to apply $A^2 + B^2 + C^2 \geq A + B + C$ for $ABC = 1$, once for $\frac{x^2}{y^2}, \frac{y^2}{z^2}, \frac{z^2}{x^2}$ and once for the reciprocals, which holds since $A+B+C \geq 3\sqrt[3]{ABC} = 3$ and $A^2 + B^2 + C^2 \geq \frac{(A+B+C)^2}{3} \geq A+B+C$ by mean inequalities. (Alternatively, apply $a_1^2 + a_2^2 + a_3^2 + a_4^2 + a_5^2 + a_6^2 \geq a_1 + a_2 + a_3 + a_4 + a_5 + a_6$ for $a_1a_2a_3a_4a_5a_6 = 1$ or just sum the AM-GM $\frac{x^2}{y^2} + \frac{y^2}{z^2} \geq 2\frac{x}{z}$ and its five analogues.)

We first note that by Muirhead, $a^2+b^2+c^2\ge a+b+c = \sum_{cyc} a^{4/3}b^{1/3}c^{1/3}$ as $(2,0,0)\succ (4/3,1/3,1/3)$. Similarly, we have $a^2b^2+b^2c^2+c^2a^2\ge ab+bc+ca = \sum_{cyc}a^{5/3}b^{5/3}c^{2/3}$, as $(2,2,0)\succ (5/3,5/3,2/3)$. Furthermore, note that by AM-GM, $a+b+c\ge 3$ and $ab+bc+ca\ge 3$. Also, if we let $u=x+y+z$ and $v=ab+bc+ca$, observe that $(a^2+b^2+c^2)(a^2b^2+b^2c^2+c^2a^2) \ge uv$. Now, rewrite the given inequality as \[2(a^2+b^2+c^2)(a^2b^2+b^2c^2+c^2a^2) \ge 3(u+v).\]From our work above, it suffices to prove that $2uv\ge 3(u+v)$, or $(u-\tfrac{3}{2})(v-\tfrac{3}{2})\ge \tfrac{9}{4}$. But since $u,v\ge 3$, we have $u-\tfrac{3}{2}, v-\tfrac{3}{2}\ge \tfrac{3}{2}$, and we are done. $\blacksquare$