suppose that $x \geq y \geq z \geq 0$. from $x^2+y^2+z^2=2(xy+yz+zx)$ we deduce that $x=(a+b)^2,y=a^2,z=b^2$ where $a \geq b \geq 0$. after some algebraic work, the inequality is equivalent $4a^6+4b^6+12a^5b+12ab^5 \geq 3 a^4b^2 +3a^2b^4+26a^3b^3$. this inequality has been proven by AM-GM

one way is using uvw method. actually i solved this problem in the exam with uvw method but it took me two hours to solve it.but the good thing was that the second problem was very easy and i solved it in half an hour

acupofmath wrote: let $ x,y,z \ge 0 $ three real nubers such that : $ x^2+y^2+z^2=2(xy+yz+zx) $ prove that $ \dfrac{x+y+z}{3} \ge \sqrt[3]{2xyz} $

hint

see here : http://www.artofproblemsolving.com/Forum/viewtopic.php?f=151&t=193555

Suppose that $z = \min (x,y,z)$. By AM-GM we have : $\frac{x+y+z}{3}=\frac{\frac{x+y-z}{2} + \frac{x+y-z}{2} + 2z}{3} \geq \sqrt[3]{2z \frac{(x+y-z)^2}{4}} = \sqrt[3]{2xyz}$

see here : http://www.artofproblemsolving.com/Forum/viewtopic.php?p=3481074#p3481074

We may perform the substitution $x = \frac{a}{bc}, y = \frac{b}{ca}, z = \frac{c}{ab}$, with $a+b+c = 0$, and where 2 of $a, b, $ and $c$ are negative. The desired inequality now becomes $a^2+b^2+c^2 \ge 3\sqrt[3]{2a^2b^2c^2} \Leftrightarrow a^2+b^2+(a+b)^2 \ge 3\sqrt{2a^2b^2(a+b)^2}$, which is easily seen to be true through smoothing, as substituting $a' = \frac{a+b}{2}$, $b' = \frac{a+b}{2}$, will yield a quantity whose value is inbetween the left and right hand sides of the inequality.

WLOG, let $x \leq y \leq z$. We also let $z = 4a$. Then, our constraint becomes $x^2 + y^2 + 16a^2 = 2(xy + 4ay + 4ax)$ or $16a^2 - 8a(x+y) + (x-y)^2 = 0$. By the quadratic formula, this becomes: \[ a = \frac{8(x+y) \pm \sqrt{64(x+y)^2 - 64(x-y)^2}}{32} = \frac{(x+y) \pm 2\sqrt{xy}}{4} = \left(\frac{\sqrt{x} \pm \sqrt{y}}{2}\right)^2 \] As $y \geq x$, this becomes \[ \sqrt{a} = \frac{\sqrt{y} \pm \sqrt{x}}{2} \] Now, only one of the above cases is possible as $\sqrt{a} = \frac{\sqrt{z}}{2} \geq \frac{\sqrt{y}}{2}$ Thus, $\sqrt{a} = \frac{\sqrt{y} + \sqrt{x}}{2}$ Now, by AM-GM, \[ \frac{\sqrt{x} + \sqrt{y}}{2} \geq \sqrt[4]{xy} \Leftrightarrow a^2 \geq xy \Leftrightarrow a \geq \sqrt[3]{xya} \] Also by AM-GM, \[ \frac{x + y + a}{3} \geq \sqrt[3]{xya} \] Adding these last two inequalities, we have: \[ \frac{x+y+4a}{3} \geq 2\sqrt[3]{xya} = \sqrt[3]{2xy(4a)} \Leftrightarrow \frac{x+y+z}{3} \geq \sqrt[3]{2xyz} \] Equality holds in AM-GM when $x=y=a$ or, equivalently, when $z = 4x = 4y$.

We use function,let $ x+y+z=1,$ so $ xy+yz+zx=\frac{1}{4} $ $ xyz=r $,$ f(t)=t^3-t^2+\frac{t}{4}-r $ $ f'(t)=3t^2-2t+\frac{1}{4}=0 $ $ t=\frac{1}{2},\frac{1}{6} $ $ f(\frac{1}{6})\geq0,f(\frac{1}{2})\leq0 $ so $ r\leq\frac{1}{54} $ over!

Let $ x\ge y \ge z $ and $ y=a^2 ,z=b^2 $ then we get that $ x=(a+b)^2 $.After some algebra we must prove that $ 4(a^2+b^2+ab)^3 \ge ((a+b)ab)^2 $.But it's very easy ,because we have $ 4(a^2+ab+b^2) \ge 3(a+b)^2 $.

Also here : http://www.artofproblemsolving.com/Forum/viewtopic.php?f=51&t=587737

Is it allowed to use UVW method on Iran MO? Because it's easily proved with it

mahanmath wrote: Suppose that $z = \min (x,y,z)$. By AM-GM we have : $\frac{x+y+z}{3}=\frac{\frac{x+y-z}{2} + \frac{x+y-z}{2} + 2z}{3} \geq \sqrt[3]{2z \frac{(x+y-z)^2}{4}} = \sqrt[3]{2xyz}$ very nice proof

of course

Let $z \ge y \ge x$ , then we have this lemma by some calculating: Lemma : $x^2+y^2+z^2=2(xy+yz+zx)$ iff $\sqrt{x} + \sqrt{y} = \sqrt{z}$ So it's enough to prove that: $\dfrac{x+y+{(\sqrt{x}+\sqrt{y})}^2}{3} \ge \sqrt[3]{2xy{(\sqrt{x}+\sqrt{y})}^2}$ Let $a = \frac{\sqrt{x}}{\sqrt{y}}$ Then by simplification we need to prove that $\dfrac{a+\frac{1}{a}+1}{3} \ge \sqrt[3]{\frac{a}{4} + \frac{1}{4a}}$ Now let $t=a + \frac{1}{a}$ Then the assertion becomes : $t +1 \ge \sqrt[3]{\frac{t}{4}}$ And $t+1 = t + \frac{1}{2} + \frac{1}{2} \ge \sqrt[3]{\frac{t}{4}}$ by AM_GM

AmirAlison wrote: Is it allowed to use UVW method on Iran MO? Because it's easily proved with it I think if you'll prove this method, you may use it.

let $x+y+z=p$ ,$xy + yz +zx=q$ ,$xyz=r$ .we want to prove that if $p^2 =4q$ ,then $p^3 > 54r *$.now ,let's make $p ,q$ fix. now ,it's enough to prove $*$ for the maximum value of $r$. but in this case we should have at least 2 of $x ,y ,z$ are equal. the rest is easy...

acupofmath wrote: Let $ x,y,z $ be three non-negative real numbers such that \[x^2+y^2+z^2=2(xy+yz+zx). \]Prove that \[\dfrac{x+y+z}{3} \ge \sqrt[3]{2xyz}.\] https://artofproblemsolving.com/community/c6h193555p1062660 h

X2+y2+z2-2xy-2yz-2zx=(x+y-z)2-4xy>0 so (x+y-z)2>4xy if x>y>z then we have x+y-z>2(xy)1/2 Or x+y+z>2(xy)1/2+2z=xy1/2+xy1/2+2z>3(2xyz)1/2 (AM-GM)