Cần chứng minh: $$\sum\limits_{}^{}\dfrac{x^2}{x^4+yz}\le \dfrac{3}{2}$$

Thật vậy theo Cauchy có:

$$\dfrac{x^2}{x^4+yz}\le \dfrac{x^2}{2x^2\sqrt{yz}}=\dfrac{1}{\sqrt{yz}}\le \dfrac{1}{4}\left(\dfrac{1}{y}+\dfrac{1}{z}\right)\\\to \sum\limits_{}^{}\dfrac{x^2}{x^4+yz}\le \dfrac{1}{2}\left(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}\right)\\\le \dfrac{1}{2}.\dfrac{xy+yz+xz}{xyz}\\\le \dfrac{1}{2}.\dfrac{x^2+y^2+z^2}{xyz}\\\le \dfrac{3}{2}$$

Dấu "=" xảy ra khi và chỉ khi: $x=y=z=1$

Đặt 

\begin{array}{l}A=\dfrac{x^2}{x^4+yz}+\dfrac{y^2}{y^4+xz}+\dfrac{z^2}{z^4+xy}\end{array}

Áp dụng BĐT AM-GM, ta có:

\begin{array}{l}\dfrac{x^2}{x^4+yz}\le \dfrac{x^2}{2\sqrt{x^4yz}}=\dfrac{1}{2\sqrt{yz}}\end{array}

Tương tự:

\begin{array}{l}\dfrac{y^2}{y^4+xz}\le \dfrac{1}{2\sqrt{xz}}; \dfrac{z^2}{x^4+zy}\le\dfrac{1}{2\sqrt{xy}}\end{array}

\begin{array}{l}→A\le\dfrac{1}{2}\left(\dfrac{1}{\sqrt{yz}}+\dfrac{1}{\sqrt{zx}}+\dfrac{1}{\sqrt{xy}}\right)\end{array}

Mà lại có:

\begin{array}{l}x^2+y^2+z^2\ge xy+yz+zx\end{array}

\begin{array}{l}→A\le\dfrac{1}{2}\left(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}\right)=\dfrac{1}{2} . \dfrac{x^2+y^2+z^2}{xyz}=\dfrac{3}{2}\end{array}

Dấu "=" xảy ra khi \(x=y=z=1.\)