Problem
Source: IMO ShortList 1998, algebra problem 3
Tags: inequalities, rearrangement inequality, 3-variable inequality, IMO Shortlist, algebra, High School Olympiads
22.10.2004 17:52
Please post your solutions. This is just a solution template to write up your solutions in a nice way and formatted in LaTeX. But maybe your solution is so well written that this is not required finally. For more information and instructions regarding the ISL/ILL problems please look here: introduction for the IMO ShortList/LongList project and regardingsolutions :
25.10.2004 10:45
Amplify the first, second, and third fraction by $x,y,z$ respectively. The LHS becomes $\sum_{cyc}\frac{x^4}{x(1+y)(1+z)}\ge \frac{(x^2+y^2+z^2)^2}{x+y+z+2(xy+yz+zx)+3}\ge \frac{(x^2+y^2+z^2)^2}{4(x^2+y^2+z^2)}\ge\frac 34$. I used the inequalities: $x^2+y^2+z^2\ge xy+yz+zx$, $x^2+y^2+z^2\ge 3$ and $x^2+y^2+z^2\ge \frac{(x+y+z)^2}3\ge x+y+z$.
20.04.2007 21:36
$\sum(\frac{x^{3}}{(y+1)(z+1)}+\frac{y+1}{8}+\frac{z+1}{8})\geq \frac{3}{4}\sum x$, therefore $\text{LHS}\geq-\frac{3}{4}+\frac{x+y+z}{2}\geq-\frac{3}{4}+\frac{3\sqrt[3]{xyz}}{2}=-\frac{3}{4}+\frac{3}{2}=\frac{3}{4}=\text{RHS}\; .$
03.11.2012 04:51
we have $3-1-1=1$ then $\sum \frac{x^3}{(1+y)(1+z)}\geq \frac{(x+y+z)^3}{(3+x+y+z)^2}\geq \frac{(x+y+z)^3}{4(x+y+z)^2}\geq \frac{3}{4}$ http://www.artofproblemsolving.com/Forum/viewtopic.php?f=51&t=498928
03.11.2012 05:02
Let $x,y$ and $z$ be positive real numbers such that $xyz=1$. Prove that \[\frac{x^{n}}{(\lambda + y)(\lambda+ z)}+\frac{y^{n}}{(\lambda + z)(\lambda + x)}+\frac{z^{n}}{(\lambda + x)(\lambda + y)} \geq \frac{3}{(\lambda+1)^2}. \]($n\ge1,\lambda\ge1$) http://www.artofproblemsolving.com/Forum/viewtopic.php?f=52&p=2838066 Let $x,y,z$ be three positive real numbers such that $xyz=1$. Show that: \[ \displaystyle \frac{x}{(1+x)(1+y)}+\frac{y}{(1+y)(1+z)}+ \frac{z}{(1+z)(1+x)} \geq \frac{3}{4}. \] http://www.artofproblemsolving.com/Forum/viewtopic.php?f=52&p=521094
21.01.2013 16:03
sqing wrote: Let $x,y$ and $z$ be positive real numbers such that $xyz=1$. Prove that \[\frac{x^{n}}{(\lambda + y)(\lambda+ z)}+\frac{y^{n}}{(\lambda + z)(\lambda + x)}+\frac{z^{n}}{(\lambda + x)(\lambda + y)} \geq \frac{3}{(\lambda+1)^2}. \]($n\ge1,\lambda\ge1$) We have: \[\sum\limits_{cyc} {\frac{{{x^n}}}{{(\lambda + y)(\lambda + z)}}} \ge \frac{{{{(x + y + z)}^n}}}{{{3^{n - 2}}\left[ {\sum {(\lambda + y)(\lambda + z)} } \right]}} \quad \quad\quad\quad(1)\] But $ {\sum {(\lambda + y)(\lambda + z)} } =3 \lambda ^2 +2\lambda(x+y+z)+(xy+yz+zx)$ and using $(1)$ it suffices to show that \[{(\lambda + 1)^2}{(x + y + z)^n} \ge {3^{n - 1}}\left[ {3{\lambda ^2} + 2\lambda (x + y + z) + (xy + yz + zx)} \right]\quad \quad\quad\quad(2)\] Observe that by AM-GM using $xyz=1$ we have $ x + y + z \ge 3 \cdot \sqrt[3]{{xyz}} \Leftrightarrow x + y + z \ge 3$. Also we have: \[{\lambda ^2}{(x + y + z)^n} \ge {3^n}{\lambda ^2}\] \[2\lambda {(x + y + z)^n} = 2\lambda {(x + y + z)^{n - 1}}(x + y + z) \ge 2\lambda \cdot {3^{n - 1}}(x + y + z)\]\[{(x + y + z)^n} = {(x + y + z)^{n - 2}}{(x + y + z)^2} \ge {3^{n - 2}} \cdot 3(xy + yz + zx) = {3^{n - 1}}(xy + yz + zx)\] Summing the above we have: \[{(x + y + z)^n}\left( {{\lambda ^2} + 2\lambda + 1} \right) \ge {3^{n - 1}}\left[ {{\lambda ^2} + 2\lambda (x + y + z) + (xy + yz + zx)} \right]\] $Q.E.D.$ We have equality only if $x=y=z=1$.
16.06.2013 16:16
orl wrote: Let $x,y$ and $z$ be positive real numbers such that $xyz=1$. Prove that \[ \frac{x^{3}}{(1 + y)(1 + z)}+\frac{y^{3}}{(1 + z)(1 + x)}+\frac{z^{3}}{(1 + x)(1 + y)} \geq \frac{3}{4}. \] $ (\frac{8x^{3}}{(1 + y)(1 + z)} + (1+y) + (1+z)) + (\frac{8y^{3}}{(1 + z)(1 + x)} + (1+z) + (1+x)) + (\frac{8z^{3}}{(1 + x)(1 + y)} + (1+x) + (1+y))\geq 3(\sqrt{8x^3}+\sqrt{8y^3}+\sqrt{8z^3})= 6(x+y+z), $ so $ \frac{8x^{3}}{(1 + y)(1 + z)} + \frac{8y^{3}}{(1 + z)(1 + x)} + \frac{8z^{3}}{(1 + x)(1 + y)} + 6 \geq 4(x+y+z) \geq 12$, from which the result follows.
24.04.2014 04:58
By Holder's inequality, \begin{align*} \left(\frac{x^{3}}{(1+y)(1+z)}+\frac{y^{3}}{(1+z)(1+x)}+\frac{z^{3}}{(1+x)(1+y)}\right)((1+y) + (1+z) + (1+x))((1+z) + (1+x) + (1+y)) \ge (x+y+z)^3 \end{align*} which is equivalent to \[ \frac{x^{3}}{(1+y)(1+z)}+\frac{y^{3}}{(1+z)(1+x)}+\frac{z^{3}}{(1+x)(1+y)} \ge \frac{(x+y+z)^3}{(3 + x + y + z)^2} \] We have that \[ \frac{(x+y+z)^3}{(3 + x + y + z)^2} = \frac{(x+y+z)^2}{(x+y+z) + 6 + \frac{9}{x+y+z}} \ge \frac{(x+y+z)^2}{(x+y+z) + 9}, \] where $ \frac{9}{x+y+z} \le 3 $ follows from $ x + y + z \ge 3(xyz)^{1/3} = 3 $. To conclude the proof we note that \[ \begin{align*} \frac{(x+y+z)^2}{(x+y+z) + 9} = \frac{(x+y+z)^2}{(x+y+z) + 9(xyz)^{1/3}} \ge \frac{(x+y+z)^2}{x+y+z + 3(x+y+z)} = \frac{(x+y+z)^2}{4(x+y+z)} = \frac{x+y+z}{4} \ge \frac{3(xyz)^{1/3}}{4} = \frac{3}{4} \end{align*} ,\] quod erat demonstrandum.
24.06.2014 20:29
this can be proven just using rearrangement inequality: \[\sum_{cyc} \frac{x^3}{(1+y)\cdot(1+z)}\geq \sum_{cyc} \frac{x^3}{(x+1)^2}\]. Then it remains to prove: \[\sum_{cyc} \frac{x^3}{(x+1)^2}\geq \frac{3}{4}\]. Using supstitution:$x=\frac{a}{b}$ we get:\[\sum_{cyc} \frac{a^3}{b(a+b)^2}\geq \sum_{cyc} \frac{a^3}{a(2a)^2}\geq \frac{1}{4}+\frac{1}{4}+\frac{1}{4}=\frac{3}{4}\].Which is true by rearrangement inequality. $Q.E.D.$
24.06.2014 21:28
hello, we set $x=a/b,y=b/c,z=c/a$ and we get $4\,{a}^{8}{c}^{4}+4\,{a}^{7}{c}^{4}b+4\,{b}^{8}{a}^{4}+4\,{b}^{7}{a}^{ 4}c+4\,{c}^{7}{b}^{4}a+4\,{c}^{8}{b}^{4}-3\,{b}^{4}{c}^{3}{a}^{5}-3\,{ b}^{5}{c}^{3}{a}^{4}-6\,{b}^{4}{c}^{4}{a}^{4}-3\,{b}^{5}{c}^{4}{a}^{3} -3\,{b}^{3}{c}^{4}{a}^{5}-3\,{b}^{3}{c}^{5}{a}^{4}-3\,{b}^{4}{c}^{5}{a }^{3} \geq 0$ and with $b=a+u,c=a+u+v$ we obtain $\left( 94\,{u}^{2}+94\,{v}^{2}+94\,uv \right) {a}^{10}+ \left( 658\,{ u}^{3}+1169\,{u}^{2}v+282\,{v}^{3}+1075\,u{v}^{2} \right) {a}^{9}+ \left( 5248\,{u}^{3}v+5757\,{u}^{2}{v}^{2}+386\,{v}^{4}+2078\,{u}^{4} +2587\,u{v}^{3} \right) {a}^{8}+ \left( 3041\,{v}^{4}u+10710\,{u}^{2}{ v}^{3}+12711\,{u}^{4}v+3914\,{u}^{5}+302\,{v}^{5}+17070\,{u}^{3}{v}^{2 } \right) {a}^{7}+ \left( 2051\,u{v}^{5}+19278\,{u}^{5}v+4886\,{u}^{6} +31164\,{u}^{4}{v}^{2}+10465\,{v}^{4}{u}^{2}+25186\,{u}^{3}{v}^{3}+140 \,{v}^{6} \right) {a}^{6}+ \left( 20313\,{v}^{4}{u}^{3}+5937\,{u}^{2}{ v}^{5}+812\,u{v}^{6}+37323\,{u}^{5}{v}^{2}+19579\,{u}^{6}v+36\,{v}^{7} +4234\,{u}^{7}+37094\,{u}^{4}{v}^{3} \right) {a}^{5}+ \left( 35755\,{u }^{5}{v}^{3}+4\,{v}^{8}+1960\,{u}^{2}{v}^{6}+30165\,{u}^{6}{v}^{2}+176 \,u{v}^{7}+24389\,{u}^{4}{v}^{4}+13660\,{u}^{7}v+9505\,{u}^{3}{v}^{5}+ 2582\,{u}^{8} \right) {a}^{4}+ \left( 16\,{v}^{8}u+6513\,{u}^{8}v+2520 \,{v}^{6}{u}^{3}+22694\,{u}^{6}{v}^{3}+18602\,{u}^{5}{v}^{4}+1094\,{u} ^{9}+16416\,{u}^{7}{v}^{2}+9097\,{u}^{4}{v}^{5}+344\,{v}^{7}{u}^{2} \right) {a}^{3}+ \left( 24\,{v}^{8}{u}^{2}+1820\,{v}^{6}{u}^{4}+9184 \,{u}^{7}{v}^{3}+2040\,{u}^{9}v+5208\,{u}^{5}{v}^{5}+5796\,{u}^{8}{v}^ {2}+308\,{u}^{10}+336\,{v}^{7}{u}^{3}+8820\,{v}^{4}{u}^{6} \right) {a} ^{2}+ \left( 1204\,{u}^{9}{v}^{2}+164\,{v}^{7}{u}^{4}+1652\,{u}^{6}{v} ^{5}+2380\,{v}^{4}{u}^{7}+380\,{u}^{10}v+16\,{v}^{8}{u}^{3}+2156\,{u}^ {8}{v}^{3}+700\,{u}^{5}{v}^{6}+52\,{u}^{11} \right) a+224\,{u}^{9}{v}^ {3}+4\,{u}^{12}+224\,{u}^{7}{v}^{5}+112\,{u}^{6}{v}^{6}+32\,{u}^{5}{v} ^{7}+4\,{v}^{8}{u}^{4}+32\,{u}^{11}v+112\,{u}^{10}{v}^{2}+280\,{u}^{8} \geq 0$ Sonnhard.
24.06.2014 22:11
Dr Sonnhard Graubner wrote: hello, we set $x=a/b,y=b/c,z=c/a$ and we get $4\,{a}^{8}{c}^{4}+4\,{a}^{7}{c}^{4}b+4\,{b}^{8}{a}^{4}+4\,{b}^{7}{a}^{ 4}c+4\,{c}^{7}{b}^{4}a+4\,{c}^{8}{b}^{4}-3\,{b}^{4}{c}^{3}{a}^{5}-3\,{ b}^{5}{c}^{3}{a}^{4}-6\,{b}^{4}{c}^{4}{a}^{4}-3\,{b}^{5}{c}^{4}{a}^{3} -3\,{b}^{3}{c}^{4}{a}^{5}-3\,{b}^{3}{c}^{5}{a}^{4}-3\,{b}^{4}{c}^{5}{a }^{3} \geq 0$ and with $b=a+u,c=a+u+v$ we obtain $\left( 94\,{u}^{2}+94\,{v}^{2}+94\,uv \right) {a}^{10}+ \left( 658\,{ u}^{3}+1169\,{u}^{2}v+282\,{v}^{3}+1075\,u{v}^{2} \right) {a}^{9}+ \left( 5248\,{u}^{3}v+5757\,{u}^{2}{v}^{2}+386\,{v}^{4}+2078\,{u}^{4} +2587\,u{v}^{3} \right) {a}^{8}+ \left( 3041\,{v}^{4}u+10710\,{u}^{2}{ v}^{3}+12711\,{u}^{4}v+3914\,{u}^{5}+302\,{v}^{5}+17070\,{u}^{3}{v}^{2 } \right) {a}^{7}+ \left( 2051\,u{v}^{5}+19278\,{u}^{5}v+4886\,{u}^{6} +31164\,{u}^{4}{v}^{2}+10465\,{v}^{4}{u}^{2}+25186\,{u}^{3}{v}^{3}+140 \,{v}^{6} \right) {a}^{6}+ \left( 20313\,{v}^{4}{u}^{3}+5937\,{u}^{2}{ v}^{5}+812\,u{v}^{6}+37323\,{u}^{5}{v}^{2}+19579\,{u}^{6}v+36\,{v}^{7} +4234\,{u}^{7}+37094\,{u}^{4}{v}^{3} \right) {a}^{5}+ \left( 35755\,{u }^{5}{v}^{3}+4\,{v}^{8}+1960\,{u}^{2}{v}^{6}+30165\,{u}^{6}{v}^{2}+176 \,u{v}^{7}+24389\,{u}^{4}{v}^{4}+13660\,{u}^{7}v+9505\,{u}^{3}{v}^{5}+ 2582\,{u}^{8} \right) {a}^{4}+ \left( 16\,{v}^{8}u+6513\,{u}^{8}v+2520 \,{v}^{6}{u}^{3}+22694\,{u}^{6}{v}^{3}+18602\,{u}^{5}{v}^{4}+1094\,{u} ^{9}+16416\,{u}^{7}{v}^{2}+9097\,{u}^{4}{v}^{5}+344\,{v}^{7}{u}^{2} \right) {a}^{3}+ \left( 24\,{v}^{8}{u}^{2}+1820\,{v}^{6}{u}^{4}+9184 \,{u}^{7}{v}^{3}+2040\,{u}^{9}v+5208\,{u}^{5}{v}^{5}+5796\,{u}^{8}{v}^ {2}+308\,{u}^{10}+336\,{v}^{7}{u}^{3}+8820\,{v}^{4}{u}^{6} \right) {a} ^{2}+ \left( 1204\,{u}^{9}{v}^{2}+164\,{v}^{7}{u}^{4}+1652\,{u}^{6}{v} ^{5}+2380\,{v}^{4}{u}^{7}+380\,{u}^{10}v+16\,{v}^{8}{u}^{3}+2156\,{u}^ {8}{v}^{3}+700\,{u}^{5}{v}^{6}+52\,{u}^{11} \right) a+224\,{u}^{9}{v}^ {3}+4\,{u}^{12}+224\,{u}^{7}{v}^{5}+112\,{u}^{6}{v}^{6}+32\,{u}^{5}{v} ^{7}+4\,{v}^{8}{u}^{4}+32\,{u}^{11}v+112\,{u}^{10}{v}^{2}+280\,{u}^{8} \geq 0$ Sonnhard. Is this uvw theoreme?Could you explain it or gve me some articles about it?
24.06.2014 23:24
is this not uvw method,it's buffalo way method.
08.07.2014 02:34
orl wrote: Let $x,y$ and $z$ be positive real numbers such that $xyz=1$. Prove that \[ \frac{x^{3}}{(1 + y)(1 + z)}+\frac{y^{3}}{(1 + z)(1 + x)}+\frac{z^{3}}{(1 + x)(1 + y)} \geq \frac{3}{4}. \]
05.08.2014 08:19
product (1+x)(1+y)(1+z) at LHS, and RHS, and it tis enough to prove that sigma(x^4+x)>=3/4 (2+sigma (x+xy)) and it is easy by xyz=1. sigma(x^4)>=sigma(xy) because sigma(x^4)>=sigma(x^2yz)=sigma(x)>=sigma(xy)
25.08.2014 18:41
From Andeescu's Inequality, we pbtain that: LHS>=(x+y+z)^3/3[(1+y)(1+z) + (1+x)(1+y) + (1+z)(1+x)] So, it is enoygh to prove that: (x+y+z)^3/3[(1+y)(1+z) + (1+x)(1+y) + (1+z)(1+x)]>=3/4 => 4(x+y+z)^3>=9(xy+yz+zx+3+2x+2y+2z) But (x+y+z)^2 (x+y+z)^3=(x+y+z)(x+y+z)^2>=9(xy+yz+zx), (becayse xyz=1 => x+y+z>=3) (1) And: x+y+z>=3 => (x+y+z)^3>=27 (2) And: 2(x+y+z)^3=2(x+y+z)(x+y+z)^2>=18(x+y+z) (3) Adding (2), (3), (1), we obtain that: 4(x+y+z)^3>=9(xy+yz+zx) + 27 + 18(x+y+z) => 4(x+y+z)^3>=9(xy+yz+zx+3+2x+2y+2z)
30.10.2014 10:04
i'll conclude N.T.TUAN's 1st claim with a bit different way : after $\frac{x^3}{(1+y)(1+z)} + \frac{1+y}{8} +\frac{1+z}{8} \ge \frac{3x}{4}$ it concludes with $\sum_{cycl}\frac{x^3}{(1+y)(1+z)}+\frac{1}{4}\sum_{cycl}(1+x)\ge \sum_{cycl}\frac{3x}{4}$ => $\sum_{cycl}\frac{x^3}{(1+y)(1+z)} \ge \frac{1}{4}\sum_{cycl}(2x-1) \ge \frac{3}{4}$.Equality holds when $x=y=z=1$
30.10.2014 13:06
Another approach is to amplify by $(1+x)(1+y)(1+z)$ and then argue by $S=x+y+z$.
26.08.2015 07:57
a good approach is in secrets in inequalities pg18
01.04.2017 12:20
Let $a,b,c>0$ and $a+b+c=1$ . Prove that$$\frac{a^5}{(b+1)(c+1)}+\frac{b^5}{(c+1)(a+1)}\frac{c^5}{(a+1)(b+1)}\geq\frac{1}{144}$$
24.10.2023 20:47
Holder's gives us \[\left(\sum_{\text{cyc}} \frac{x^3}{(1+y)(1+z)} \right)\left(\sum_{\text{cyc}} (1+y) \right)\left(\sum_{\text{cyc}} (1+z) \right) \ge (x+y+z)^3\] Setting $s=x+y+z$, it suffices to prove \[\frac{s^3}{(3+s)^2} \ge \frac{3}{4}\] Simplifying, we want \[(s-3)(4s^2+9s+9) \ge 0\] so we want to prove $s = a + b + c \ge 3$, which is true by AM-GM. $\square$
21.01.2024 23:40
Legitimate solve (with a 25% ARCH hint) By Hölders Inequality $$\left (\sum_{\text{cyc}} \frac{x^{3}}{(1+y)(1+z)} \right) \left( \sum_{\text{cyc}} 1+y \right) \left( \sum_{\text{cyc}} 1+z \right) \geq (x+y+z)^{3}$$Now we can turn this into a $1$ variable inequality, let $s=x+y+z$. It suffices to show $$\frac{s^{3}}{(3+s)^{2}} \geq \frac{3}{4}$$Which is true if $s \geq 3$, which is true by AM-GM. Finishing
06.02.2024 08:11
Note that the inequality rearranges as \[\sum_{\mathrm{cyc}} x^3(1+x)\ge\frac{3}{4}\prod_{\mathrm{cyc}}(1+x).\]Make the substitution $x=a^3$, $y=b^3$, $z=c^3$. Homogenizing, the inequality reduces to showing \[4\sum_{\mathrm{cyc}}a^9(abc+a^3)\ge 3abc\prod_{\mathrm{cyc}}(abc+a^3).\]Expanding, this reduces down to showing \[4\sum_{\mathrm{cyc}}a^{10}bc + 4\sum_{\mathrm{cyc}}a^12\ge 6a^4b^4c^4 + 3\sum_{\mathrm{cyc}}a^6b^3c^3 + 3\sum_{\mathrm{cyc}}a^5b^5c^2.\]This follows from adding the following four inequalities, all of which directly follow from Murihead. \begin{align*} 3\sum_{\mathrm{cyc}}a^{10}bc &\ge\sum_{\mathrm{cyc}}a^5b^5c^2 \\ 3\sum_{\mathrm{cyc}}a^{12} &\ge\sum_{\mathrm{cyc}}a^6b^3c^3 \\ \sum_{\mathrm{cyc}}a^{10}bc &\ge 3a^4b^4c^4 \\ \sum_{\mathrm{cyc}}a^{12} &\ge 3a^4b^4c^4. \end{align*}
21.02.2024 08:08
By Holder's we get \[(\sqrt[3]{3})\left(\sum_{cyc} (1+y)(1+z) \right)^{\frac{1}{3}} \left(\sum_{cyc} \frac{x^3}{(1+y)(1+z)} \right)^{\frac{1}{3}} \geq (x+y+z)\]Cubing, it remains to prove that $4(x+y+z)^3 \geq 9\sum_{cyc} (1+y)(1+z)$. Simplifying we get that we need \[12\sum_{sym}x^2y + 4 \sum_{cyc} x^3 \geq 18(x+y+z)+ 9(xy+xz+yz)\]Homogonizing we get from Muirhead that \[\sum_{sym} x^{\frac{5}{3}}y^{\frac{2}{3}}z^{\frac{-1}{3}} \geq \sum_{sym} xy\]\[\sum_{sym} x^{\frac{8}{3}}y^{\frac{-1}{3}}z^{\frac{-1}{3}} \geq \sum_{sym} xy\]\[\sum_{sym} x^{\frac{4}{3}}y^{\frac{1}{3}}z^{\frac{-2}{3}} \geq \sum_{sym} x\]\[\sum_{sym} x^{\frac{7}{3}}y^{\frac{-2}{3}}z^{\frac{-2}{3}} \geq \sum_{sym} x\]Combining all of these we can obviously see the result
10.03.2024 19:15
Multiplying both sides by $4(1+x)(1+y)(1+z)$ means we want to prove $$4x^3+4x^4+4y^3+4y^4+4z^3+4z^4\ge 3+3x+3y+3z+3xy+3yz+3zx+3xyz.$$We use $xyz=1$ to homogenize the desired inequality into $$4(\sum_{\text{cyc}}x^{\frac{10}3}y^{\frac13}z^{\frac13}+\sum_{\text{cyc}}x^4)\ge 3(2x^\frac43y^\frac43z^\frac43+\sum_{\text{cyc}}x^2yz+\sum_{\text{cyc}}x^{\frac53}y^{\frac53}z^\frac23).$$Since $x^4+x^4+y^4+z^4\ge 4x^2yz$, $\sum_{\text{cyc}}x^4\ge \sum_{\text{cyc}}x^2yz$. Since $5x^4+5y^4+2z^4\ge 12x^\frac53y^\frac53z^\frac23$, $\sum_{\text{cyc}}x^4\ge \sum_{\text{cyc}}x^\frac53y^\frac53z^\frac23$. Therefore, we want to prove that $$2\sum_{\text{cyc}}x^\frac{10}3y^\frac13z^\frac13\ge 3x^\frac43y^\frac43z^\frac43+\sum_{\text{cyc}}x^{\frac53}y^{\frac53}z^\frac23.$$Since $\sum_{\text{cyc}}x^\frac{10}3y^\frac13z^\frac13\ge 3x^\frac43y^\frac43z^\frac43$ by AM-GM, we just need to prove that $$\sum_{\text{cyc}}x^\frac{10}3y^\frac13z^\frac13\ge \sum_{\text{cyc}}x^{\frac53}y^{\frac53}z^\frac23$$. By AM-GM, $4x^\frac{10}3y^\frac13z^\frac13+4x^\frac{1}3y^\frac{10}3z^\frac13+x^\frac{1}3y^\frac{1}3z^\frac{10}3\ge 9x^\frac53y^\frac53z^\frac23$ and summing cyclically proves the desired result.
28.04.2024 21:36
oops apparently i had the holders idea but applied it wrong... i am not marking as solve ig also titu $x/x$ trick... should probably know to like do something with it and use simple bounds at least xd oops Let $S$ be the sum. Notice \[S\cdot \sum_{\text{cyc}}(1+y)\cdot \sum_{\text{cyc}}(1+z)\ge (x+y+z)^3\]by Holder which implies \[S\ge \frac{(x+y+z)^3}{(3+x+y+z)^2}.\] Let $s=x+y+z$. Notice \[\frac{s^3}{s^2+6s+9}\ge \frac{3}{4}\iff (s-3)(4s^2+9s+9)\ge 0\]and since \[s^3\ge 27xyz=27\implies s\ge 3\]we are done. $\blacksquare$
28.04.2024 21:37
also not sure if we can get rearrangement to work here, but \[\frac{x^3}{(1+x)^2}=\frac{x^2}{\tfrac{1}{x}+2+x}\]and applying Titu yields something like $(x-1)^2(4x+5)\ge 0$.
27.07.2024 07:50
We can use Muirhead's inequality too
04.09.2024 04:17
Basically the same as several solutions above. We use Hölder's inequality as follows: \[\left(\sum_\text{cyc} \frac{x^3}{(1 + y)(1 + z)}\right)(3 + x + y + z)^2 \geq (x + y + z)^3\]\[ \implies \sum_\text{cyc} \frac{x^3}{(1 + y)(1 + z)} \geq \frac{(x + y + z)^3}{(3 + x + y + z)^2}.\]Also, we have $x + y + z \geq 3\sqrt[3]{xyz} = 3$ by AM-GM. Setting $s = x + y + z$, we wish to prove that \[\frac{s^3}{(3 + s)^2} \geq \frac34\]for $s \geq 3$. But this reduces to $4s^3 - 3s^2 - 18s - 27 \geq 0$, or $(4s^2 + 9s + 9)(s - 3) \geq 0$, which is trivial since $s \geq 3$. $\square$
08.09.2024 05:08
Holders gives us $$(\frac{x^{3}}{(1 + y)(1 + z)}+\frac{y^{3}}{(1 + z)(1 + x)}+\frac{z^{3}}{(1 + x)(1 + y)})^{\frac13} \cdot (3+x+y+z)^{\frac13} \cdot (3+x+y+z)^{\frac13} \geq x^3+y^3+z^3$$so proving $$\frac{(x+y+z)^3}{(3+x+y+z)(3+x+y+z)} \geq \frac34$$is sufficient, which reduces to $x+y+z \geq 3$ which is true after a straight AM-GM so we are done.
12.10.2024 00:52
Very nice question . We have $x,y,z \in \mathbb{R^{+}}$ and $xyz=1$. Claim: $\boxed{\frac{x^{3}}{(1 + y)(1 + z)}+\frac{y^{3}}{(1 + z)(1 + x)}+\frac{z^{3}}{(1 + x)(1 + y)} \geq \frac{3}{4}}$ Proof: We can homogenize the above inequality by substituting $(x,y,z)=\left(\dfrac{a}{b},\dfrac{b}{c},\dfrac{c}{a}\right)$ where $a,b,c \in \mathbb{R^{+}}$.Thus the claim can be restated as: $$ \sum_{cyc} \dfrac{\dfrac{a^{3}}{b^{3}}}{\left(1+\dfrac{b}{c}\right)\left(1+\dfrac{c}{a}\right)} \geq \frac{3}{4}$$$$ \iff \sum_{cyc} \dfrac{a^{4}c}{b^{3}(c+b)(c+a)} \geq \frac{3}{4} $$$$ \iff \sum_{cyc} a^{7}c^{4}(a+b) \geq \frac{3}{4}a^{3}b^{3}c^{3}(a+b)(b+c)(c+a) $$$$ \iff 4(a^{8}c^{4}+b^{8}a^{4}+c^{8}b^{4}+a^{7}c^{4}b+b^{7}a^{4}c+c^{7}b^{4}a) \geq 3(a^{5}b^{4}c^{3}+a^{5}c^{4}b^{3}+b^{5}a^{4}c^{3}+b^{5}c^{4}a^{3}+c^{5}a^{4}b^{3}+c^{5}b^{4}a^{3})+6a^{4}b^{4}c^{4} $$In order to prove this, consider the following inequalities which are a direct application of AM-GM inequality: $$ a^{8}c^{4}+b^{7}a^{4}c+a^{4}b^{4}c^{4}+b^{5}a^{4}c^{3} \geq 4a^{5}b^{4}c^{3} \cdots (1) $$$$ b^{8}a^{4}+c^{7}b^{4}a+a^{4}b^{4}c^{4}+c^{5}b^{4}a^{3} \geq 4b^{5}c^{4}a^{3} \cdots (2) $$$$ c^{8}b^{4}+a^{7}c^{4}b+a^{4}b^{4}c^{4}+a^{5}c^{4}b^{3} \geq 4c^{5}a^{4}b^{3} \cdots (3)$$$$ a^{8}c^{4}+a^{4}b^{4}c^{4}+b^{5}c^{4}a^{3} \geq 3a^{5}b^{3}c^{4} \cdots (4) $$$$ b^{8}a^{4}+a^{4}b^{4}c^{4}+c^{5}a^{4}b^{3} \geq 3b^{5}c^{3}a^{4} \cdots (5)$$$$ c^{8}b^{4}+a^{4}b^{4}c^{4}+a^{5}b^{4}c^{3} \geq 3c^{5}a^{3}b^{4} \cdots (6)$$$$ a^{8}c^{4}+b^{8}a^{4}+c^{8}b^{4} \geq 3a^{4}b^{4}c^{4} \cdots (7)$$$$ a^{7}c^{4}b+b^{7}a^{4}c+c^{7}b^{4}a \geq 3a^{4}b^{4}c^{4} \cdots(8)$$Now by applying Weighted AM-GM inequality we will get the following inequalities: $$ 5a^{3}b^{6}+2b^{3}c^{6}+2c^{3}a^{6} \geq 9a^{3}b^{4}c^{2} $$$$ 5b^{3}c^{6}+2c^{3}a^{6}+2a^{3}b^{6} \geq 9b^{3}c^{4}a^{2} $$$$ 5c^{3}a^{6}+2a^{3}b^{6}+2b^{3}c^{6} \geq 9c^{3}a^{4}b^{2} $$Adding these $3$ inequalities and dividing by $9$ in both the sides we shall get: $$a^{3}b^{6}+b^{3}c^{6}+c^{3}a^{6} \geq a^{3}b^{4}c^{2}+a^{2}b^{3}c^{4}+a^{4}b^{2}c^{3} $$Now multiplying both the sides by $abc$ we have, $$b^{7}a^{4}c+c^{7}b^{4}a+a^{7}c^{4}b \geq a^{5}b^{3}c^{4}+b^{5}c^{3}a^{4}+c^{5}a^{3}b^{4} \cdots (9) $$Now finally doing $(1)+(2)+(3)+(4)+(5)+(6)+2(7)+2(8)+(9)$ we get, $$4(a^{8}c^{4}+b^{8}a^{4}+c^{8}b^{4}+a^{7}c^{4}b+b^{7}a^{4}c+c^{7}b^{4}a) \geq 3(a^{5}b^{4}c^{3}+a^{5}c^{4}b^{3}+b^{5}a^{4}c^{3}+b^{5}c^{4}a^{3}+c^{5}a^{4}b^{3}+c^{5}b^{4}a^{3})+6a^{4}b^{4}c^{4}$$as desired. $\blacksquare$
12.10.2024 14:54
I keep forgetting Holder exists, maybe I need to actually practice it. If this is a fakesolve someone tell me. Note by Cauchy-Schwarz: \begin{align*} \sum_{\textrm{cyc}}\frac{x^3}{(1+y)(1+z)}&=\sum_{\textrm{cyc}}\frac{x^4}{x(1+y)(1+z)}\\ &\ge\frac{(x^2+y^2+z^2)^2}{\sum_{\textrm{cyc}}x(1+y)(1+z)}\\ &\ge\frac{\frac{1}{9}(x+y+z)^4}{\sum_{\textrm{cyc}}(x+xy+xz+xyz)}\\ &=\frac{\frac{1}{9}(x+y+z)^4}{\left(\sum_{\textrm{cyc}}x\right)+2\left(\sum_{\textrm{cyc}}xy\right)+3}\\ &\ge\frac{1}{9}\frac{(x+y+z)^4}{\left(\sum_{\textrm{cyc}}x\right)+\frac{2}{3}\left(\sum_{\textrm{cyc}}x\right)^2+3} \end{align*}So we want to show that if $s=x+y+z$ then $\frac{4}{27}s^4\ge s+\frac{2}{3}s^2+3$ after rearranging. Note however by AM-GM that $s=x+y+z\ge3\sqrt[3]{xyz}=3$, so if $P(s)=\frac{1}{27}(4s^4-18s^2-27s-81)$ then it suffices to show that $P(s)\ge0$ for all $s\ge3$. Note however that $P(3)=0$ which can be checked, and $P’(s)=\frac{1}{27}(16s^3-36s-27)$, which as $s^3=(s^2)s\ge9s$, $P’(s)\ge\frac{1}{27}(16(9s)-36s-27)=\frac{1}{27}(144s-36s-27)=\frac{1}{27}(108s-27)\ge\frac{1}{27}(297)=11>0$, which implies that $P(s)$ is strictly increasing over the interval $[3,\infty)$. Hence the inequality is true. Ext1 techniques carrying here
27.10.2024 13:46
27.10.2024 14:01
Sketch: After AM-GM on denominator, we have $(3, -1/2, -1/2) > (2/3, 2/3, 2/3)$ so we're done.
24.12.2024 00:44
what the bad Write $x=a^3$, $y=b^3$, and $z=c^3$, so that $abc=1$. We find \[\sum(a^{12}+a^9abc)=\sum(a^12+a^9)\ge \frac{3}{4}(1+a^3)(1+b^3)(1+c^3)=\frac{3}{4}abc(abc+a^3)(abc+b^3)(abc+c^3)\]where this inequality is homogeneous. It suffices to prove \[4(a^{12}+b^{12}+c^{12})+4(a^{10}bc+ab^{10}c+abc^{10})\ge 6a^4b^4c^4+3(a^6b^3c^3+a^3b^6c^3+a^3b^3c^6)+3(a^5b^5c^2+a^5b^2c^5+a^2b^5c^5)\]and all sums can be reduced to a symmetric sum in $a$, $b$, and $c$, where everything on the left majorizes everything on the right, and furthermore plugging in $a=b=c=1$ yields $24\ge 24$, i.e. coefficients match up. So we are done. $\blacksquare$ yay muirhead