Power mean $\implies (\frac{a^\frac{3}{2}+b^\frac{3}{2}+c^\frac{3}{2}}{3})^2\ge (\frac{a+b+c}{3})^3=\frac{1}{27}\implies (a^\frac{3}{2}+b^\frac{3}{2}+c^\frac{3}{2})^2\ge\frac{1}{3}$ C-S(T2) $\implies \left( \frac{a^3}{x^2+2y^2} + \frac{b^3}{y^2+2z^2} + \frac{c^3}{z^2+2x^2} \right)\ge \frac{(a^\frac{3}{2}+b^\frac{3}{2}+c^\frac{3}{2})^2}{3(x^2+y^2+z^2)}\ge \frac{1}{9(x^2+y^2+z^2)}$ (P.S.Please check,I have did somethink silly yesterday )

socrates wrote: Let $x,y,z$ and $a,b,c$ be positive real numbers with $a+b+c=1$. Prove that $$\displaystyle \left(x^2+y^2+z^2\right) \left( \frac{a^3}{x^2+2y^2} + \frac{b^3}{y^2+2z^2} + \frac{c^3}{z^2+2x^2} \right) ~\ge~~ \frac19$$ Using Holder we have $$\displaystyle \left(x^2+y^2+z^2\right) \left( \frac{a^3}{x^2+2y^2} + \frac{b^3}{y^2+2z^2} + \frac{c^3}{z^2+2x^2} \right) \geq (x^2+y^2+z^2) \left(\frac{1}{9(x^2+y^2+z^2)}\right)=\frac{1}{9}.$$

socrates wrote: Let $x,y,z$ and $a,b,c$ be positive real numbers with $a+b+c=1$. Prove that $$\displaystyle \left(x^2+y^2+z^2\right) \left( \frac{a^3}{x^2+2y^2} + \frac{b^3}{y^2+2z^2} + \frac{c^3}{z^2+2x^2} \right) ~\ge~~ \frac19$$ 2017 Mediterranean Mathematics Olympiad . Problem 4 Proof of Zhangyunhua: $\left(x^2+y^2+z^2\right) \left( \frac{a^3}{x^2+2y^2} + \frac{b^3}{y^2+2z^2} + \frac{c^3}{z^2+2x^2} \right)$ $=\frac{1}{3}\left((x^2+2y^2)+(y^2+2z^2)+(z^2+2x^2)\right) \left( \frac{a^3}{x^2+2y^2} + \frac{b^3}{y^2+2z^2} + \frac{c^3}{z^2+2x^2} \right)$ $\geq \frac{1}{3} \left(x^{\frac{3}{2}}+y^{\frac{3}{2}}+z^{\frac{3}{2}}\right)^2\geq \frac19$

sqing wrote: socrates wrote: Let $x,y,z$ and $a,b,c$ be positive real numbers with $a+b+c=1$. Prove that $$\displaystyle \left(x^2+y^2+z^2\right) \left( \frac{a^3}{x^2+2y^2} + \frac{b^3}{y^2+2z^2} + \frac{c^3}{z^2+2x^2} \right) ~\ge~~ \frac19$$ 2017 Mediterranean Mathematics Olympiad . Problem 4 Proof of Zhangyunhua: $\left(x^2+y^2+z^2\right) \left( \frac{a^3}{x^2+2y^2} + \frac{b^3}{y^2+2z^2} + \frac{c^3}{z^2+2x^2} \right)$ $=\frac{1}{3}\left((x^2+2y^2)+(y^2+2z^2)+(z^2+2x^2)\right) \left( \frac{a^3}{x^2+2y^2} + \frac{b^3}{y^2+2z^2} + \frac{c^3}{z^2+2x^2} \right)$ $\geq \frac{1}{3} \left(x^{\frac{3}{2}}+y^{\frac{3}{2}}+z^{\frac{3}{2}}\right)^2\geq \frac19$ $\geq \frac{1}{3} \left(x^{\frac{3}{2}}+y^{\frac{3}{2}}+z^{\frac{3}{2}}\right)^2\geq \frac19$ can you explain this part pls? thanks

AlgebraFC wrote:

what is holder's inequality?

You can use Google search, but anyway see here to learn many classical inequalities (Holder's inequality is also here) : https://brilliant.org/wiki/classical-inequalities/

Absolutely trivial. By Holder, we have $$\begin{align*} ((x^2+2y^2)+(y^2+2z^2)+(z^2+2x^2))(1+1+1)\left( \frac{a^3}{x^2+2y^2} + \frac{b^3}{y^2+2z^2} + \frac{c^3}{z^2+2x^2} \right)\geq (a+b+c)^3. \end{align*}$$ Rearrange and we get desired.

Exactly same as above, and I also think this is trivial. By Holder we have $$\left(\sum_{\text{cyc}} \frac{a^3}{x^2 + 2y^2}\right) \cdot \left(\sum_{\text{cyc}} x^2 + 2y^2\right) \cdot \left(\sum_{\text{cyc}} 1\right) \geq 1$$ . Rearranging this gives the result.

$$\text{ Notice that }$$ $$\sum_{cyc}x^2\cdot\sum_{cyc}\frac{a^3}{x^2+2y^2}\overset{\text{Generalized T2'S}}{\ge}\sum_{cyc}x^2\cdot\frac{\left(\sum_{cyc}a\right)^3}{9\sum_{cyc}x^2}=\frac{\sum_{cyc}x^2}{9\sum_{cyc}x^2}=\frac{1}{9}$$ $$\blacksquare.$$