Now, it seems a little bit more organized if we let \[ \{\begin{array}{cc} a=x^2+y^2+z^2 \\ \\ b=xy+yz+zx \end{array}\] the inequality turns out into $ (a-b)(a+3b) \ge 0$ and we are done!

Let $ x^2+y^2+z^2=m$ and $ xy+yz+zx=n$ , then the inequality become $ \frac{m+n}{2}\le\sqrt{m+2n}\sqrt{\frac{m}{3}}$ $ \left(\frac{m+n}{2}\right)^2\le\frac{m^2+2mn}{3}$ $ 3(m^2+n^2+2mn)\le4(m^2+2mn)$ $ 0\le m^2-3n^2+2mn$ $ 0\le (m-n)(m+3n)$ , which it true cause $ m\ge n$ by $ AM-GM$ and $ m+3n\ge0$

Let $ x+y+z=p$ and $ xy+yz+zx=q$. Then the inequality turns into $ \frac{p^2-q}{2}\le p \sqrt{\frac{p^2-2q}{3}}$. The last is equivalent to $ p^4-2p^2q-3q^2\ge 0$. Examining the last as quadratic inequality towards $ p^2$ we get roots $ -q$ and $ 3q$. So since we want to prove the inequality we need to prove $ p^2 \in \left(-\infty,-q\right)$ and $ p^2 \in \left(3q,+\infty\right)$. Since $ p^2\ge 0$ we only need to prove that $ p^2 \in \left(3q,+\infty\right)$ which is obvious since $ p^2\ge 3q$ is equivalent after expanding to $ \left(x-y\right)^2+\left(y-z\right)^2+\left(z-x\right)^2\ge 0$.

Since given inequality is homogeneous, W.L.O.G. we may assume that $ x^2 + y^2 + z^2 = 3$. Then it is enough to prove that: \[ 2(x + y + z)\geq3 + xy + yz + zx = \frac {3 + (x + y + z)^2}{2},\] or \[ (\lambda - 1)(\lambda - 3)\leq0,\] where $ \lambda = x + y + z$; which is true since $ x^2 + y^2 + z^2 = 3$.

April wrote: Let $ x$, $ y$ and $ z$ be non negative numbers. Prove that \[ \frac{x^2+y^2+z^2+xy+yz+zx}{6}\le \frac{x+y+z}{3}\cdot\sqrt{\frac{x^2+y^2+z^2}{3}}\] Strengthening Let $ x$, $ y$ and $ z$ be non negative numbers. Prove that \[ x^2+y^2+z^2+xy+yz+zx\le (x+y+z)\cdot\sqrt{x^2+y^2+z^2+\frac{xy+yz+zx}{3}}\]

sqing wrote: April wrote: Let $ x$, $ y$ and $ z$ be non negative numbers. Prove that \[ \frac{x^2+y^2+z^2+xy+yz+zx}{6}\le \frac{x+y+z}{3}\cdot\sqrt{\frac{x^2+y^2+z^2}{3}}\] Strengthening Let $ x$, $ y$ and $ z$ be non negative numbers. Prove that \[ x^2+y^2+z^2+xy+yz+zx\le (x+y+z)\cdot\sqrt{x^2+y^2+z^2+\frac{xy+yz+zx}{3}}\] let $ x+y+z=3a, xy+yz+xz=3b^2 $ and $a\geq b$ $\iff 9a^2-6b^3+3b^2\leq 3a\sqrt{9a^2-6b^2+\frac{3b^2}{3}} \iff$ $(3a^2-b^2)^2\leq a^2(9a^2-5b^2) \iff $ $b^4\leq a^2b^2 \iff b\leq a$ this is true.

KöMaL 2006 Let $ x$, $ y$ be positive real numbers. Prove that \[ \frac{x^2+y^2+xy}{3}\le \frac{x+y}{2}\cdot\sqrt{\frac{x^2+y^2}{2}}\]

sqing wrote: KöMaL 2006 Let $ x$, $ y$ be positive real numbers. Prove that \[ \frac{x^2+y^2+xy}{3}\le \frac{x+y}{2}\cdot\sqrt{\frac{x^2+y^2}{2}}\] Let $ x+y=2a, xy=b^2 $ we have $a\geq b$ $\iff \frac{(4a^2-2b^2+b^2)}{3}\leq\frac{2a}{2}\sqrt{\frac{4a^2-2b^2}{2}} $ $\iff (4a^2-b^2)^2\leq 9a^2(2a^2-b^2) $ $\iff 2a^4\geq a^2b^2+b^4 $ this is true.

April wrote: Let $ x$, $ y$ and $ z$ be non negative numbers. Prove that \[ \frac{x^2+y^2+z^2+xy+yz+zx}{6}\le \frac{x+y+z}{3}\cdot\sqrt{\frac{x^2+y^2+z^2}{3}}\] By uvw method, we may check the case $z=0$ or $y=z$ case1:$z=0$ The inequality is equivalent to $x^4+y^4+2(x^3y+xy^3)\geq x^2y^2$ This holds by AM-GM inequality case2:$y=z$ The inequality is equivalent to $x^4+5y^4+4x^3y\geq 6x^2y^2+4xy^3$ By AM-GM, $x^4+y^4\geq 2x^2y^2$ By Rearrangement, $y^3+x^3\geq x^2y+xy^2\rightarrow 4y^4+4x^3y\geq 4x^2y^2+4xy^3$ So we're done

sqing wrote: Strengthening Let $ x$, $ y$ and $ z$ be non negative numbers. Prove that \[ x^2+y^2+z^2+xy+yz+zx\le (x+y+z)\cdot\sqrt{x^2+y^2+z^2+\frac{xy+yz+zx}{3}}\] By uvw method, we may check the case $z=0$ or $y=z$. case1:$z=0$ The inequality is equivalent to $x^3y+xy^3\geq x^2y^2$ This holds by AM-GM inequality case2:$y=z$ The inequality is equivalent to $\frac{2}{3}x^3y+\frac{1}{3}y^4\geq x^2y^2$ This holds by AM-GM.So we're done

very nice Takeya.O

sqing wrote: Strengthening Let $ x$, $ y$ and $ z$ be non negative numbers. Prove that \[ x^2+y^2+z^2+xy+yz+zx\le (x+y+z)\cdot\sqrt{x^2+y^2+z^2+\frac{xy+yz+zx}{3}}\] Notice that $ (x+y+z)^2 = x^2 + y^2 + z^2 + 2(xy + yz + zx) $ Also, by AM-GM, $ x^2 + y^2 + z^2 \geq xy + yz + zx $ Let $ x^2 + y^2 + z^2 = a $ and $xy + yz + zx = b $ Now we know that $ (x+y+z)^2 = a +2b $ and we have $ a \geq b $ Now we may convert the whole original inequality into $ a,b $, which is: $ \sqrt {a+2b} \cdot \sqrt {a+\frac{b}{3}} \geq a + b $ Since both sides are non-negative we may square the inequality $ \Rightarrow (a+2b)(a+\frac{b}{3}) \geq a^2 + 2ab + b^2 $ $ \Rightarrow a^2 + \frac{ab}{3} + 2ab + \frac{2b^2}{3} \geq a^2 + 2ab + b^2 $ $ \Rightarrow a + 2b \geq 3b $ $ \Rightarrow a \geq b $ Which we've already stated to be true by AM-GM

Notice we need to prove $\dfrac{x+y+z}{3}\sqrt{\dfrac{x^2+y^2+z^2}{3}}=\sqrt{\dfrac{4(x+y+z)^2(x^2+y^2+z^2)}{108}} \geq \sqrt{\dfrac{3(x^2+y^2+z^2+xy+yz+zx)^2}{108}}=\dfrac{x^2+y^2+z^2+xy+yz+zx}{6}$ So we are left with \[4(x+y+z)^2(x^2+y^2+z^2) \geq 3(x^2+y^2+z^2+xy+yz+zx)^2\]which is true since \[4(x+y+z)^2(x^2+y^2+z^2)=2(x+y+z)^2*2(x^2+y^2+z^2) \geq 3(x^2+y^2+z^2+xy+yz+zx)(x^2+y^2+z^2+xy+yz+zx)=3(x^2+y^2+z^2+xy+yz+zx)^2\]By AM-GM and the case of equality is WLOG $x=y, z=0$, and $x=y=z$.

Let $s=x+y+z$ and $p=xy+xz+yz$. The inequality becomes $$s^2-p\leq 2s\sqrt{(s^2-2p)/3}$$$$3(s^2-p)^2\leq 4s^2(s^2-2p).$$This expands out to $$s^4-2s^2p-3p^2\geq 0.$$Note that $p\leq \frac{1}{3}s^2$, so $$s^4=\frac{1}{3}s^2(2s^2+s^2)\geq p(2s^2+3p)\rightarrow s^4-2s^2p-3p^2\geq 0,$$so we are done.

Sketch: Let $x^2 + y^2 + z^2 = a, xy + yz + zx = b$. Then the inequality is equivalent to $$(a-b)(a+3b) \ge 0.$$$a -b \ge 0$ is known and $a+3b \ge 0$ is because $a, b \ge 0$. $\square$