$\frac {ab}{a + b} + \frac {bc} {b + c} + \frac {ca} {c + a} = \frac{2}{3}(\frac{1}{ac+bc}+\frac{1}{ab+ac}+\frac{1}{bc+ab}) \geq \frac{3}{ab+bc+ca}$ $\frac{3(a+b+c)}{3(a^3+b^3+c^3)}\leq \frac{3(a+b+c)}{a^3+b^3+c^3+6abc} \leq \frac{3(a+b+c)}{(a+b+c)(ab+bc+ca)}=\frac{3}{ab+bc+ca}$

sqing wrote: Let $a, b, c $ be positive real numbers such that $abc = \frac {2} {3}. $ Prove that: $$\frac {ab}{a + b} + \frac {bc} {b + c} + \frac {ca} {c + a} \geqslant \frac {a+b+c} {a^3+b ^ 3 + c ^ 3}.$$ Solution. Noticing that $x^3+y^3\ge xy(x+y)$ holds for any $x,y\in\mathbb{R}^+$, we obtain \begin{align*} \frac {ab}{a + b} + \frac {bc} {b + c} + \frac {ca} {c + a}=&\frac {(ab)^2}{ab(a + b)} + \frac {(bc)^2} {bc(b + c)} + \frac {(ca)^2} {ca(c + a)}\\ \left(\because x^3+y^3\ge xy(x+y)\right)\geqslant&\frac {(ab)^2}{a^3 + b^3} + \frac {(bc)^2} {b^3 + c^3} + \frac {(ca)^2} {c^3 + a^3}\\ (\because\text{Cauchy-Schwarz's Inequality})\geqslant&\frac {(ab+bc+ca)^2} {2\left(a^3+b ^ 3 + c ^ 3\right)}\\ \left(\because (x+y+z)^2\ge3(xy+yz+zx)\right)\geqslant&\frac {3abc(a+b+c)} {2\left(a^3+b ^ 3 + c ^ 3\right)}\\ \left(\because abc = \frac {2} {3}\right)=& \frac {a+b+c} {a^3+b ^ 3 + c ^ 3}. \end{align*}As required. $\blacksquare$

The inequality we want to show is nothing but $$ (a^3+b^3+c^3)\left(\sum_{cyc}\frac{ab}{a+b}\right)\geqslant a+b+c. $$First, observe that, $$ \frac{ab}{a+b}=\frac{abc}{ac+bc}=\frac{2}{3}\cdot\frac{1}{ac+bc} \implies \sum_{cyc}\frac{ab}{a+b}=\frac23\sum_{cyc}\frac{1}{ac+bc}\geqslant \frac{3}{ab+bc+ca}. $$Hence, the inequality boils down proving: $$ 3(a^3+b^3+c^3)\geqslant (ab+bc+ca)(a+b+c). $$Now, by Cauchy-Schwarz, we have $(a^3+b^3+c^3)(a+b+c)\geqslant (a^2+b^2+c^2)^2$, which, by another application of Cauchy-Schwarz with $3(a^2+b^2+c^2)\geqslant (a+b+c)^2$, yields that, $$ (a^3+b^3+c^3)(a+b+c)\geqslant (a+b+c)^2(a^2+b^2+c^2)/3\geqslant 3(a+b+c)^2(ab+bc+ca) $$which implies, $$ a^3+b^3+c^3\geqslant (a+b+c)(ab+bc+ca)/3, $$using the well-known inequality $a^2+b^2+c^2\geqslant ab+bc+ca$. Edit: Indeed, I made a mistake in the earlier version. Thanks to ytChen for the alert. The proof above is hopefully fixed.

grupyorum wrote: The inequality we want to show is nothing but $$ (a^3+b^3+c^3)\left(\sum_{cyc}\frac{ab}{a+b}\right)\geqslant a+b+c. $$First, observe that, $$ \frac{ab}{a+b}=\frac{abc}{ac+bc}=\frac{2}{3}\cdot\frac{1}{ac+bc} \implies \sum_{cyc}\frac{ab}{a+b}=\frac23\sum_{cyc}\frac{1}{ac+bc}\geqslant \frac{3}{ab+bc+ca}. $$Hence, the inequality boils down proving: $$ (a^3+b^3+c^3)\geqslant 3(ab+bc+ca)(a+b+c). $$Now, by Cauchy-Schwarz, we have $(a^3+b^3+c^3)(a+b+c)\geqslant (a^2+b^2+c^2)^2$, which, by another application of Cauchy-Schwarz with $3(a^2+b^2+c^2)\geqslant (a+b+c)^2$, yields that, $$ (a^3+b^3+c^3)(a+b+c)\geqslant 3(a+b+c)^2(a^2+b^2+c^2)\geqslant 3(a+b+c)^2(ab+bc+ca) $$which implies, $$ a^3+b^3+c^3\geqslant 3(a+b+c)(ab+bc+ca), $$using the well-known inequality $a^2+b^2+c^2\geqslant ab+bc+ca$. The fourth line from the bottom of the solution seems problematic, it should have been $ 3(a^3+b^3+c^3)(a+b+c)\geqslant (a+b+c)^2(a^2+b^2+c^2)$ instead of $ (a^3+b^3+c^3)(a+b+c)\geqslant 3(a+b+c)^2(a^2+b^2+c^2)$.

This problem was proposed by Dimitar Trenevski from FYR Macedonia.

RagvaloD wrote: $\frac {ab}{a + b} + \frac {bc} {b + c} + \frac {ca} {c + a} = \frac{2}{3}(\frac{1}{ac+bc}+\frac{1}{ab+ac}+\frac{1}{bc+ab}) \geq \frac{3}{ab+bc+ca}$ $\frac{3(a+b+c)}{3(a^3+b^3+c^3)}\leq \frac{3(a+b+c)}{a^3+b^3+c^3+6abc} \leq \frac{3(a+b+c)}{(a+b+c)(ab+bc+ca)}=\frac{3}{ab+bc+ca}$ excuse me,l can't obtain the last inequality

enliar wrote: RagvaloD wrote: $\frac {ab}{a + b} + \frac {bc} {b + c} + \frac {ca} {c + a} = \frac{2}{3}(\frac{1}{ac+bc}+\frac{1}{ab+ac}+\frac{1}{bc+ab}) \geq \frac{3}{ab+bc+ca}$ $\frac{3(a+b+c)}{3(a^3+b^3+c^3)}\leq \frac{3(a+b+c)}{a^3+b^3+c^3+6abc} \leq \frac{3(a+b+c)}{(a+b+c)(ab+bc+ca)}=\frac{3}{ab+bc+ca}$ excuse me,l can't obtain the last inequality Which inequality ?

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

Let $a, b, c $ be positive real numbers such that $a+b+c=3. $ Prove that: $$ \frac{ab}{4bc+b+c}+\frac{bc}{4ca+c+a}+\frac{ca}{4ab+a+b}\ge\frac{abc}{2}.$$(Florin Rotaru)

Nice inequality! Didn't take me long to crack it From condition of problem we firstly get that $$\sum_{cyc}\frac{ab}{a + b} = \frac{2}{3}\sum_{cyc}\frac{1}{ac + bc} (1)$$Now by Cauchy-Schwarz we get that $$\frac{2}{3}\sum_{cyc}\frac{1}{ac + bc}\geq \frac{2}{3}\frac{(1 + 1 + 1)^2}{2(ab + bc + ca)} = \frac{3}{ab + bc + ac} (2)$$Claim: $3(\sum_{cyc}a^3)\geq(\sum_{cyc} ab)(\sum_{cyc}a)$ If we prove this we get solution, so let's get to work. Again by Cauchy-Schwarz we have $$\sum_{cyc}a^3\sum_{cyc}a\geq(\sum_{cyc}a^2)^2 (3)$$And again by Cauchy-Schwarz we have $$3(\sum_{cyc}a^2)\geq (\sum_{cyc}a)^2 (4)$$Now from $(4)$ and $(3)$ we get that $$(\sum_{cyc}a^3)(\sum_{cyc}a)\geq\frac{(\sum_{cyc}a^2)(\sum_{cyc}a)^2}{3}\geq\frac{(\sum_{cyc}a)^2(\sum_{cyc}ab)}{3} (5)$$From $(5)$ we get what we wanted, hence we are done! Comment: In last step of $(5)$ I used well known inequality $$\sum_{cyc}a^2\geq\sum_{cyc}ab$$

enliar wrote: excuse me,l can't obtain the last inequality Directly comes from Muirhead. $$3(a^3+b^3+c^3) \ge (a+b+c)(ab+bc+ca)$$

We use the standard $pqr$ notations. Our inequality becomes $$\frac{3q^2+2p}{3pq-2}\geq\frac{p}{p^3-3pq+2}$$$$3p^3q^2-9pq^3+6q^2+2p^4-6p^2c+4p\geq 3p^2q-2p$$$$3p^3q^2+6q^2+2p^4+6p\geq 9p^2q+9pq^3$$A clear adaptation of Schur's inequality gives $$2p^3+9r\geq 7pq \Leftrightarrow 2p^3+6\geq 7pq$$which rewrites equivalently as $$2p^3q^2+6q^2\geq 7pq^3$$and respectively $$2p^4+6p\geq 7p^2q$$Using the two inequalities previously highlighted we only have to prove that $$p^3q^2\geq 2p^2q+2pq^3$$From known inequalities we have $$pq\geq 9r=6\Leftrightarrow \frac{p^3q^2}{3}\geq 2p^2q$$$$p^2\geq 3q\Leftrightarrow \frac{2p^3q^2}{3}\geq 2pq^3$$which together finish the proof.

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

sqing wrote: Let $a, b, c $ be positive real numbers such that $a+b+c=3. $ Prove that: $$ \frac{ab}{4bc+b+c}+\frac{bc}{4ca+c+a}+\frac{ca}{4ab+a+b}\ge\frac{abc}{2}.$$(Florin Rotaru) By Cauchy-Schwarz's inequality: $LHS\geq \frac{(\sum ab)^{2}}{15abc+\sum ab^{2}}\geq \frac{9abc}{14+abc+\sum ab^{2}}$ It suffices to prove that $ab^{2}+bc^{2}+ca^{2}+abc\leq 4$ WLOG, assume that $b$ is between $a$ and $c$. $\Rightarrow a(a-b)(b-c)\geq 0$ $\Rightarrow ab^{2}+bc^{2}+ca^{2}+abc\leq b(a+c)^{2}=4.b.\frac{a+c}{2}.\frac{a+c}{2}\leq 4(\frac{b+\frac{a+c}{2}+\frac{a+c}{2}}{3})^{3}=4$ We have $Q.E.D$. $\square$

VMF-er wrote: By Cauchy-Schwarz's inequality: $LHS\geq \frac{(\sum ab)^{2}}{15abc+\sum ab^{2}}\geq \frac{9abc}{14+abc+\sum ab^{2}}$ I think there is some mistake in the denominator. How will it be $15abc$ ? I think it will be something like $4abc(a+b+c)+3abc+(\sum ab)^{2}$

please correct me if wrong* abc=2/3 a minimum =(2/3)^1/3 b minimum =(2/3)^1/3 c minimum =(2/3)^1/3 then on substituting each of the values we get the inequality to be satisifying the equation

Solved with Xbenx First we will use the fact that $abc = \frac {2} {3}. $ to homogenize the inequality: $\frac {ab}{a+b}=\frac {2} {3c(a+b)}$...(1) Now putting ...(1) to $\frac {ab}{a + b} + \frac {bc} {b + c} + \frac {ca} {c + a} \geqslant \frac {a+b+c} {a^3+b ^ 3 + c ^ 3}.$ we get $$\frac{2}{3}\sum_{cyc}\frac{1}{c(a + b)}\geq \frac{(a+b+c)} {(a^3 + b^3 + a^3)} $$$$\sum_{cyc}\frac{1}{c(a + b)}\geq \frac{3(a+b+c)} {2(a^3 + b^3 + a^3)} $$ now let us apply Cauchy-Schwarz's Inequality: $(a^3(b+c)+b^3(a+c)+c^3(a+b))*(\sum_{cyc}\frac{1}{c(a + b)})\geq (a+b+c)^2$ From this we are left to prove that $2(a+b+c)(a^3+b^3+c^3)=3(a^3(b+c)+b^3(a+c)+c^3(a+b))$ whic is equivalent to $2a^4+2b^4+2c^4\geq a^3b+a^3c+b^3a+b^3c+c^3a+c^3b$ Now we can finish from $AM-GM$ From $AM-GM$ we have $a^3+a^3+a^3+b^3\geq 4a^3b$ similarly , we get that $2a^4+2b^4+2c^4\geq a^3b+a^3c+b^3a+b^3c+c^3a+c^3b$.So we are done $\blacksquare$

p/7449427768 Let $a,b,c> 0$ and $a^2+b^2+c^2=3.$ Prove that $$\frac{1+bc}{a+bc}+\frac{1+ca}{b+ca}+\frac{1+ab}{c+ab}\geq a+b+c.$$

sqing wrote: Let $a, b, c $ be positive real numbers such that $a+b+c=3. $ Prove that: $$ \frac{ab}{4bc+b+c}+\frac{bc}{4ca+c+a}+\frac{ca}{4ab+a+b}\ge\frac{abc}{2}.$$(Florin Rotaru) Thanks.

From C-S we get $$\frac{ab}{a+b}+\frac{bc}{b+c}+\frac{ca}{c+a}=\frac{2}{3}\left(\frac{1}{c(a+b)}+\frac{1}{a(b+c)}+\frac{1}{b(c+a)}\right)\geq\frac{2}{3}\cdot\frac{9}{2(ab+bc+ca)}=\frac{3}{ab+bc+ca}$$So it is sufficient to prove $\frac{3}{ab+bc+ca}\geq\frac{a+b+c}{a^3+b^3+c^3}\iff 3(a^3+b^3+c^3)\geq(a+b+c)(ab+bc+ca)$ $a^3+b^3+c^3\geq3abc$ from AM-GM $\sum_{sym} a^3\geq \sum_{sym} a^2b$ from Murhead Adding those two inequalities we get $3(a^3+b^3+c^3)\geq(a+b+c)(ab+bc+ca)$ and we are done

One liner We can rewrite the LHS as \[\sum_{\text{cyc}} \frac{a^2b^2}{a^2b+ab^2} \ge \frac{(ab+bc+ca)^2}{\sum_{\text{cyc}} (a^2b+ab^2)} \ge \frac{3(ab \cdot bc + bc \cdot ca + ca \cdot ab)}{\sum_{\text{cyc}} (a^3+b^3)} = \frac{3abc}{2} \cdot \left(\frac{a+b+c}{a^3+b^3+c^3}\right).~\blacksquare\]

\[\sum_{\text{cyc}} \frac{a^2b^2}{a^2b+ab^2} \ge \frac{(ab+bc+ca)^2}{\sum_{\text{sym}} ab^2} \ge \frac{3(ab \cdot bc + bc \cdot ca + ca \cdot ab)}{(a^3+b^3)+(b^3+c^3)+(c^3+a^3)} = \frac{3abc}{2} \cdot \frac{a+b+c}{a^3+b^3+c^3}.~\blacksquare\]= ?????

Homogenizing and multiplying by $a^3+b^3+c^3$ this is equivalent to $\sum_{\text{sym}} \frac{a^2}b+\sum_{\text{cyc}} \frac{a^2}{b+c} \ge \frac 32(a+b+c)+\sum_{\text{cyc}}\frac{ab}c.$ By Muirhead's we have $\sum_{\text{sym}} \frac{a^2}b \ge a+b+c+\sum_{\text{cyc}}\frac{ab}c,$ and by Titu's lemma we have $\sum_{\text{cyc}}\frac{a^2}{b+c}\ge\frac{(a+b+c)^2}{2(a+b+c)}=\frac 12(a+b+c)$ so summing these finishes.

We have \[\sum_{cyc}\frac{ab}{a+b}=\frac 23 \sum_{cyc}\frac{1}{(a+b)c} \geq \frac 23 \left(\frac{9}{2(ab+ac+bc)}\right)=\frac{3}{ab+ac+bc}\]so it suffices to show $3(a^3+b^3+c^3)\geq (a+b+c)(ab+ac+bc)$ which is true by Muirhead as $(3,0,0)\succ(2,1,0)$ and AM-GM on $a^3+b^3+c^3\geq 3abc$.

We use $abc=\tfrac{2}{3}$ to note that \[\sum_{cyc}\frac{ab}{a+b} = \frac{2}{3} \sum_{cyc} \frac{1}{c(a+b)}.\]Then, by Titu, we have \[\frac{2}{3}\sum_{cyc}\frac{1}{c(a+b)} = \frac{2}{3}\cdot \frac{9}{2(ab+bc+ca)} = \frac{3}{ab+bc+ca}.\]Thus, it suffices to prove that $\tfrac{3}{ab+bc+ca}\ge \tfrac{a+b+c}{a^3+b^3+c^3}$, or that \[3(a^3+b^3+c^3)\ge 3abc+ \sum_{sym}a^2b.\]But this is true, as $2(a^3+b^3+c^3)\ge \sum_{sym}a^2b$ by Muirhead and $a^3+b^3+c^3\ge 3abc$ by AM-GM, so we are done. $\blacksquare$

sqing wrote: Let $a, b, c $ be positive real numbers such that $abc = \frac {2} {3}. $ Prove that: $$\frac {ab}{a + b} + \frac {bc} {b + c} + \frac {ca} {c + a} \geqslant \frac {a+b+c} {a^3+b ^ 3 + c ^ 3}.$$ https://artofproblemsolving.com/community/c6h2267997p17621979

By Titu's Lemma, \begin{align*} \sum_{\text{cyc}} \dfrac{ab}{a+b} = \dfrac{2}{3} \cdot \sum_{\text{cyc}} \dfrac{1}{ac+bc} \geq \dfrac{2}{3} \cdot \dfrac{9}{2ab+2bc+2ca} \geq \dfrac{3}{ab+bc+ca}. \end{align*}Therefore, it suffices to show $3\left(a^3 + b^3 + c^3\right) \geq \left(ab+bc+ca\right)\left(a+b+c\right)$. This is equivalent to show \begin{align*} \sum_{\text{cyc}} 3a^3 \geq \sum_{\text{sym}} a^2b + 3abc. \end{align*}By AM-GM, note that \begin{align*} 2 \cdot \sum_{\text{cyc}} a^3 \geq \sum_{\text{sym}} a^2b \quad \text{and} \quad \sum_{\text{cyc}} a^3 \geq 3abc, \end{align*}by Muirhead's and AM-GM. Thus, summing yields the desired result.

Since $abc = \frac{2}{3}$, it suffices to prove that \[\frac{2}{3}\sum_{\text{cyc}} \frac{1}{ac+bc} \ge \frac{a+b+c}{a^3+b^3+c^3}\]Then, by Titu's inequality, we have \[\frac{2}{3}\sum_{\text{cyc}} \frac{1}{ac+bc} \ge \frac{3}{(ab+bc+ca)}\]So it suffices to prove that \[\frac{3}{ab+bc+ca} \ge \frac{a+b+c}{a^3+b^3+c^3}\]\[\Longleftrightarrow 3(a^3+b^3+c^3) \ge (a+b+c)(ab+bc+ca)\]\[\Longleftrightarrow 3 \sum_{\text{sym}} a^3 \ge 2 \sum_{\text{sym}} a^2b + \sum_{\text{sym}} abc\]which is true by Muirhead.

sqing wrote: Let $a, b, c $ be positive real numbers such that $abc = \frac {2} {3}. $ Prove that: $$\frac {ab}{a + b} + \frac {bc} {b + c} + \frac {ca} {c + a} \geqslant \frac {a+b+c} {a^3+b ^ 3 + c ^ 3}.$$ Nice problem

$$\frac{ab}{a+b}+\frac{bc}{b+c}+\frac{ca}{c+a}=\frac{2}{3}\sum_{cyc}\frac{1}{ac+bc}\overset{\text{C-S}}{\geq} \frac{3}{ab+bc+ca}\overset{\text{?}}{\geq}\frac{a+b+c}{a^3+b^3+c^3}$$$\implies3(a^3+b^3+c^3) \overset{\text{?}}{\geq} (a+b+c)(ab+bc+ca)$ By Cauchy-Schwarz, $(a^3+b^3+c^3)(a+b+c)\geq(a^2+b^2+c^2)^2$ Again by Cauchy-Schwarz, $3(a^2+b^2+c^2)\geq (a+b+c)^2$ And well-known, $a^2+b^2+c^2 \geq ab+bc+ca$ $$\implies 3(a^3+b^3+c^3)(a+b+c)\geq 3(a^2+b^2+c^2)^2 \geq (a+b+c)^2(ab+bc+ca)$$$$\implies 3(a^3+b^3+c^3)\geq (a+b+c)(ab+bc+ca)$$Done.

Firstly, note that \[\frac {ab}{a + b} + \frac {bc} {b + c} + \frac {ca} {c + a} = \sum \frac{1}{\frac 1a + \frac 1b} \stackrel{\rm Titu}{\ge} \frac{9}{\frac 2a + \frac 2b + \frac 2c}.\]Therefore, it suffices to prove (after homogenisation) that $$\frac{9}{\frac 2a + \frac 2b + \frac 2c} \ge \frac{(a+b+c)(abc)}{a^3+b^3+c^3} \times \frac 32.$$But this simplifies after expanding to $$\sum_{sym}a^3b^0c^0 + \frac 12 \times \sum_{sym}a^3b^0c^0\ge \sum_{sym} a^2b^1c^0 + \frac 12 \times \sum_{sym}a^1b^1c^1.$$which is true by Muirhead. $\square$

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Thank you.