Ta có $\dfrac{1}{2}.(\dfrac{1}{\sqrt{1+y^2}}+\dfrac{1}{ \sqrt{1+z^2}})^2$ \leq $\dfrac{1}{1+y^2}+\dfrac{1}{1+z^2} $ \leq $\dfrac{2}{1+yz} $
\Rightarrow $\dfrac{1}{\sqrt{1+y^2}}+\dfrac{1}{\sqrt{1+z^2}} $ \leq $\dfrac{2}{\sqrt{1+yz}}$
mà $\dfrac{1}{\sqrt{1+x^2}} $ \leq $\dfrac{\sqrt{2}}{x+1}$
\Rightarrow ta cần c/m $\dfrac{2}{\sqrt{1+yz}}$+$\dfrac{\sqrt{2}}{x+1}$ \leq $\dfrac{3}{\sqrt{2}}$
\Leftrightarrow $\dfrac{2\sqrt{x(x+1)}}{x+1}$+$\dfrac{\sqrt{2}}{x+1}$ \leq $\dfrac{3}{\sqrt{2}}$
\Leftrightarrow $\dfrac{-(\sqrt{x+1}-\sqrt{2x})^2}{\sqrt{2}(x+1)}$ \leq 0 (luôn đúng)
\Rightarrow dpcm