mavropnevma wrote: Prove that if $x$, $y$, $z$ are positive real numbers and $xyz = 1$ then \[\frac{x^3}{x^2+y}+\frac{y^3}{y^2+z}+\frac{z^3}{z^2+x}\geq \dfrac {3} {2}.\] Very old trick. $\sum_{cyc}\frac{x^3}{x^2+y}\geq\frac{3}{2}\Leftrightarrow\sum_{cyc}\left(\frac{x^3}{x^2+y}-x\right)\geq\frac{3}{2}-x-y-z\Leftrightarrow$ $\Leftrightarrow\sum_{cyc}\frac{-xy}{x^2+y}\geq\frac{3}{2}-x-y-z$ and $\frac{-xy}{x^2+y}\geq\frac{-xy}{2x\sqrt y}$.

mavropnevma wrote: Prove that if $x$, $y$, $z$ are positive real numbers and $xyz = 1$ then \[\frac{x^3}{x^2+y}+\frac{y^3}{y^2+z}+\frac{z^3}{z^2+x}\geq \dfrac {3} {2}.\] A. Golovanov I will present another proof beside the arqady's one. Since $xyz=1,$ we may put $x=\frac{a}{b},$ $y=\frac{c}{a}$ and $z=\frac{a}{b},$ where $a,\,b,\,c$ are positive real numbers. The inequality can be written as \[\frac{a^4}{b(a^3+b^2c)}+\frac{b^4}{c(b^3+c^2a)}+\frac{c^4}{a(c^3+a^2b)} \ge \frac{3}{2}.\] Now, using the Cauchy-Schwarz inequality, we have \[\left[\sum \frac{a^4}{b(a^3+b^2c)}\right] \left( \sum \frac{a^3+b^2c}{b}\right) \ge \left(\sum \frac{a^2}{b}\right)^2.\] Therefore, it suffices to prove that \[2\left(\frac{a^2}{b}+\frac{b^2}{c}+\frac{c^2}{a}\right)^2 \ge 3\left(\frac{a^3+b^2c}{b}+\frac{b^3+c^2a}{c}+\frac{c^3+a^2b}{a}\right).\] This inequality can be written as \[\begin{aligned} 2\left(\frac{a^4}{b^2}+\frac{b^4}{c^2}+\frac{c^4}{a^2}\right) &+4\left(\frac{a^2b}{c}+\frac{b^2c}{a}+\frac{c^2a}{b}\right ) \ge \\ &\ge 3\left(\frac{a^3}{b}+\frac{b^3}{c}+\frac{c^3}{a}\right) +3(ab+bc+ca).\end{aligned}\] On the other hand, we can see that \[\frac{a^2b}{c}+\frac{b^2c}{a}+\frac{c^2a}{b}=\frac{a^2b^2}{bc}+\frac{b^2c^2}{ca}+\frac{c^2a^2}{ab} \ge ab+bc+ca.\] Using this estimation, we come up with a new inequality: \[2\left(\frac{a^4}{b^2}+\frac{b^4}{c^2}+\frac{c^4}{a^2}\right) +(ab+bc+ca) \ge 3\left(\frac{a^3}{b}+\frac{b^3}{c}+\frac{c^3}{a}\right),\] which is true according to the AM-GM inequality: \[\frac{a^4}{b^2}+\frac{a^4}{b^2}+ab \ge 3\cdot \frac{a^3}{b}.\]

happy to see you again. btw your solution is very nice !

Dear CanVQ! You proof is very nice! But for easy inequality almost always there is nice solution. Did you see it: http://www.artofproblemsolving.com/Forum/viewtopic.php?p=3152176#p3152176 What do you think?

mavropnevma wrote: Prove that if $x$, $y$, $z$ are positive real numbers and $xyz = 1$ then \[\frac{x^3}{x^2+y}+\frac{y^3}{y^2+z}+\frac{z^3}{z^2+x}\geq \dfrac {3} {2}.\] A. Golovanov here is one more $\sum_{cyc} \frac{x^3}{x^2+y} \ge \frac {(x^2+y^2+z^2)^2} {x^3+y^3+z^3+xy+yz+zx} \ge \frac 3 2$ where the last one is equivalent to $2(x^4+y^4+z^4)+4(x^2y^2+y^2z^2+z^2x^2) \ge 3(x^3+y^3+z^3+xy+yz+zx)$ which folows from $2(x^4+y^4+z^4)+4(x^2y^2+y^2z^2+z^2x^2) \ge$ $\ge 2(x^4+y^4+z^4)+(x+y+z)+3(x^2y^2+y^2z^2+z^2x^2)=$ $=\sum_{cyc}(x^4+x^4+x)+3(x^2y^2+y^2z^2+z^2x^2) \ge $ $\ge 3(x^3+y^3+z^3+xy+yz+zx)$

mavropnevma wrote: Prove that if $x$, $y$, $z$ are positive real numbers and $xyz = 1$ then \[\frac{x^3}{x^2+y}+\frac{y^3}{y^2+z}+\frac{z^3}{z^2+x}\geq \dfrac {3} {2}.\] A. Golovanov Prove that if $x$, $y$, $z$ are positive real numbers and $xyz = 1$ then \[\frac{z^3}{x+z^2}+\frac{x^3}{x^2+y}+\frac{y^3}{z+y^2} \geq \frac{1}{2}(x+y+z)\geq \frac{3}{2}\]

And you do: \[\frac{z^3}{x+z^2}+\frac{x^3}{x^2+y}+\frac{y^3}{z+y^2} \] \[\geq \frac{x^2}{x+y}+\frac{y^2}{y+z}+\frac{z^2}{z+x}\geq \frac{1}{2}(x+y+z)\] BQ

The following inequality is also true. Let $x$, $y$, $z$ are positive numbers such that $xyz=1$. Prove that\[\frac{x^3+y}{x^2+y}+\frac{y^3+z}{y^2+z}+\frac{z^3+x}{z^2+x}\geq\frac{x^2+y}{x+y}+\frac{y^2+z}{y+z}+\frac{z^2+x}{z+x}\]\[\geq\frac{x^2+y}{1+y}+\frac{y^2+z}{1+z}+\frac{z^2+x}{1+x}\geq 3 .\] http://www.artofproblemsolving.com/Forum/viewtopic.php?f=52&t=545349

arqady wrote: mavropnevma wrote: Prove that if $x$, $y$, $z$ are positive real numbers and $xyz = 1$ then \[\frac{x^3}{x^2+y}+\frac{y^3}{y^2+z}+\frac{z^3}{z^2+x}\geq \dfrac {3} {2}.\] Very old trick. $\sum_{cyc}\frac{x^3}{x^2+y}\geq\frac{3}{2}\Leftrightarrow\sum_{cyc}\left(\frac{x^3}{x^2+y}-x\right)\geq\frac{3}{2}-x-y-z\Leftrightarrow$ $\Leftrightarrow\sum_{cyc}\frac{-xy}{x^2+y}\geq\frac{3}{2}-x-y-z$ and $\frac{-xy}{x^2+y}\geq\frac{-xy}{2x\sqrt y}$. ____________ Very nice solutions! But it can and no tricks: $ \frac{x^3}{x^2+y}=x-\frac{xy}{x^2+y}\Rightarrow\sum\frac{x^3}{x^2+y}\ge x+y+z-\frac{1}{2}\sum\sqrt{y}\Rightarrow $ $ \Rightarrow LHS\ge x+y+z-\frac{1}{2}\sqrt{3(x+y+z) }$ We used Cauchy-Schwarz to get $ \sum\sqrt{y}\le\sqrt{3(x+y+z)} $ It is enough to show that: $ x+y+z-\frac{1}{2}\sqrt{3(x+y+z)}\ge\frac{3}{2} $ With substitution $ t=\sqrt{3(x+y+z)} $, considering that,by GM-AM, we have $ x,y,z>0,xyz=1\Rightarrow x+y+z\ge 3 $ result $ t\ge 3 $ and ${ x+y+z-\frac{1}{2}\sqrt{3(x+y+z)}\ge\frac{3}{2}\Leftrightarrow\frac{t^2}{3}-\frac{t}{2}\ge\frac{3}{2}\Leftrightarrow 2t^2-3t-9\ge 0}\Leftrightarrow (t-3)(2t+3)\ge 0 $ The proof is ended!

Solution from contest If $x=\max(x,y,z)>3$ the left hand side is trivially greater the $\frac{3}{2}$ since $\frac{x^3}{x^2+y}\ge \frac{x^3}{2x^2} > \frac{3}{2}$ else re-write inequality as $\sum x - \sum \frac{xy}{x^2+y} \ge \sum x -\sum \frac{\sqrt{x}}{2}$ (by AM-GM on denominators) we are left to prove that $\sum x -\sum \frac{\sqrt{x}}{2}\ge \frac{3}{2}$ replace $x,y,z$ by $e^a, e^b,e^c)$ and apply Jensen inequality on the function $ g(x)= e^x -\frac{\sqrt{e^x}}{2}$ which is convex for $e^x > \frac{1}{64}$ so there is no case left to analyse . Since for a very small $x$ we get a big $y$ since $xyz=1$.

Tuymaada 2013 senior, Q4 refined： $x$, $y$, $z$ are positive numbers such that $xyz=1$. Prove that \[\frac{x^3}{x^2+y}+\frac{y^3}{y^2+z}+\frac{z^3}{z^2+x}\geq \frac{x^2}{x+y}+\frac{y^2}{y+z}+\frac{z^2}{z+x}.\]

$\sum \frac{x^3}{x^2+y}=\sum x-\sum \frac{xy}{x^2+y}\ge \sum x-\frac{1}{2}\sum \sqrt{x}$ $=\frac{1}{2}\sum (\sqrt{x}-1)^2+\frac{3}{4}\sum x-\frac{3}{4}\ge \frac{9}{4}\sqrt[3]{xyz}-\frac{3}{4}=\dfrac {3} {2}.$

arqady wrote: mavropnevma wrote: Prove that if $x$, $y$, $z$ are positive real numbers and $xyz = 1$ then \[\frac{x^3}{x^2+y}+\frac{y^3}{y^2+z}+\frac{z^3}{z^2+x}\geq \dfrac {3} {2}.\] Very old trick. $\sum_{cyc}\frac{x^3}{x^2+y}\geq\frac{3}{2}\Leftrightarrow\sum_{cyc}\left(\frac{x^3}{x^2+y}-x\right)\geq\frac{3}{2}-x-y-z\Leftrightarrow$ $\Leftrightarrow\sum_{cyc}\frac{-xy}{x^2+y}\geq\frac{3}{2}-x-y-z$ and $\frac{-xy}{x^2+y}\geq\frac{-xy}{2x\sqrt y}$. I am a bit confused with your last step. So you have shown that $\frac{-xy}{x^2+y}\geq \frac{-xy}{2x\sqrt{y}}=-\frac{\sqrt{y}}{2}$. But why the desired inequality follows from the last one? Thanks a lot!

Because after this it's enough to prove that: $$x+y+z-\frac{\sqrt{x}+\sqrt{y}+\sqrt{z}}{2}\geq\frac{3}{2},$$which is true by Murhead and AM-GM $$x+y+z-\frac{\sqrt{x}+\sqrt{y}+\sqrt{z}}{2}\geq \sqrt{x}+\sqrt{y}+\sqrt{z}-\frac{\sqrt{x}+\sqrt{y}+\sqrt{z}}{2}=$$$$=\frac{\sqrt{x}+\sqrt{y}+\sqrt{z}}{2}\geq\frac{3}{2}.$$

Prove that if $x$, $y$, $z$ are positive real numbers and $xyz = 1$ then $$\frac{x^3}{kx^2+y}+\frac{y^3}{ky^2+z}+\frac{z^3}{kz^2+x}\geq \dfrac {3} {k+1}$$Where $k\in N^+.$ More general Prove that if $x$, $y$, $z$ are positive real numbers and $xyz = 1$ then $$\frac{x^3}{kx^2+y}+\frac{y^3}{ky^2+z}+\frac{z^3}{kz^2+x}\geq \dfrac {3} {k+1}$$Where $k>0.$

mavropnevma wrote: Prove that if $x$, $y$, $z$ are positive real numbers and $xyz = 1$ then \[\frac{x^3}{x^2+y}+\frac{y^3}{y^2+z}+\frac{z^3}{z^2+x}\geq \dfrac {3} {2}.\] A. Golovanov Azerbaijan junior olympiad camp day 5 P2