Since $3(xy+yz+zx) \leq (x+y+z)^2=9$ $\Leftrightarrow$ $(x-y)^2+(y-z)^2+(z-x)^2 \geq 0$ we are done.

The following inequality is also true. Let $x$, $y$ and $z$ be non-negative numbers such that $x+y+z=3$. Prove that: $$xy+xz+yz\leq3-\frac{16}{81}(x-y)^2(x-z)^2(y-z)^2$$

how to prove it arqady?

By the brain

(x + y + z)^2 = x^2 + y^2 + z^2 + 2(xy + yz + zx) >= 3(xy + yz + zx) or , 3(xy + yz + zx) <= 9 or , xy + yz + zx <= 3

arqady wrote: The following inequality is also true. Let $x$, $y$ and $z$ be non-negative numbers such that $x+y+z=3$. Prove that: $$xy+xz+yz\leq3-\frac{16}{81}(x-y)^2(x-z)^2(y-z)^2$$ Note that the arqady's inequality is equivalent to $(x-y)^2+(y-z)^2+(z-x)^2 \geq \frac{32}{27}(x-y)^2(x-z)^2(y-z)^2$. WLOG, assume that $x \geq y \geq z$. First, since $(x-y)^2+(y-z)^2 \geq \frac{(z-x)^2}{2}$. Next, from A.M-G.M inequality, $3=x+y+z \geq (x-y)+2(y-z) \geq 2\sqrt{2(x-y)(y-z)}$ Therefore, $81 \geq 64(x-y)^2(y-z)^2$ Thus, $(x-y)^2+(y-z)^2+(z-x)^2 \geq \frac{3(z-x)^2}{2} \geq \frac{32}{27}(x-y)^2(y-z)^2(z-x)^2$ as desired.

arqady wrote: The following inequality is also true. Let $x$, $y$ and $z$ be non-negative numbers such that $x+y+z=3$. Prove that: $$xy+xz+yz\leq3-\frac{16}{81}(x-y)^2(x-z)^2(y-z)^2$$ We can solve by uvw method! Because $f'(w^3)\leq 0$

Akatsuki1010 wrote: We can solve by uvw method! Because $f'(w^3)\leq 0$ No! $f'(w^3)\leq 0\Leftrightarrow\sum_{cyc}(2x^3-3x^2y-3x^2z+4xyz)\geq0$, which is wrong.

The maximal $k$, for which the inequality $$xy+xz+yz\leq3-k(x-y)^2(x-z)^2(y-z)^2$$is true for all non-negatives $x$, $y$ and $z$ such that $x+y+z=3$, is $k=\frac{69+11\sqrt{33}}{648}=0.20399...$

Sorry Arqady but I calculated $f'(w^3)$. What is wrong with my solution? Let $a+b+c=3u$, $ab+bc+ca=3v^2$, $abc=w^3$. Then the inequality is equivalent to: $64v^6+64u^{3}v{3}+16w^6-48u^{2}v^{4}-96uv^{2}w^{3}+3-3v^2\geq 0$ $f'(w^3)=32w^2-96uv^2\leq 32w^2-32uv^2\leq 0$ since $u,v\geq w$. Thank you very much Arqady!

Akatsuki1010 wrote: Sorry Arqady but I calculated $f'(w^3)$. What is wrong with my solution? Let $a+b+c=3u$, $ab+bc+ca=3v^2$, $abc=w^3$. Then the inequality is equivalent to: $64v^6+64u^{3}v{3}+16w^6-48u^{2}v^{4}-96uv^{2}w^{3}+3-3v^2\geq 0$ $f'(w^3)=32w^2-96uv^2\leq 32w^2-32uv^2\leq 0$ since $u,v\geq w$. If so $f'(w^3)=32w^3-96uv^2+64u^3$.

$x^2+y^2+z^2 \geq xy+yz+zx$ $\implies (x+y+z)^2 \geq 3(xy+yz+zx)$ $\implies 3 \geq xy+yz+zx$.$\blacksquare$

Very easy problem

Let $x,y,z$ be non-negative numbers such that $x+y+z=3$. Prove that: $$xy+2yz+3zx \leq \frac{27}{4}-\frac{1}{18}(x-y)^2(y-z)^2(z-x)^2$$

kapsitis wrote: Assume that real numbers $x$, $y$ and $z$ satisfy $x + y + z = 3$. Prove that $xy + xz + yz \leq 3$. LOL can't believe that this problem could be in a contest

arqady wrote: By the brain lame