<s>[INST] How find this maximum $f(x)=\sqrt{\frac{1+x^2}{2}}+\sqrt{x}-x$ let $x\ge 0$,Find the follow maximum
$$f(x)=\sqrt{\dfrac{1+x^2}{2}}+\sqrt{x}-x$$
I find 
$$f'(x)=\dfrac{1}{2}\left(\dfrac{\sqrt{2}x}{\sqrt{x^2+1}}+\dfrac{1}{\sqrt{x}}-2\right)$$
so
$$f'(x)=0\Longrightarrow x=1$$
so I think 
$$f(x)_{max}=f(1)$$
But I this methods is ugly, and I think this have AM-GM inequality or Cauchy-Schwarz inequality to solve this problem
$$f(x)=\sqrt{\dfrac{1+x^2}{2}}+\sqrt{x}-x\le\sqrt{\dfrac{1+x^2}{2}}+\dfrac{1}{2}(1+x)-x=\sqrt{\dfrac{1+x^2}{2}}+\dfrac{1}{2}(1-x)$$
Then I can't,Thank you
 [/INST] Let us show that $f(x)+x\leqslant x+1$ for every $x$. This holds true if and only if
$$
\sqrt{\frac{x+\frac1x}2}+1\leqslant\sqrt{x}+\frac1{\sqrt{x}}.
$$
Let $u=\sqrt{x}+\frac1{\sqrt{x}}$, then $u\geqslant2$ for every $x\geqslant0$ and $x+\frac1x=u^2-2$ hence it is enough to show that $g(u)\leqslant0$ for every $u\geqslant2$, where
$$
g(u)=\sqrt{\frac{u^2-2}2}+1-u.
$$
Now, $g(2)=0$ and
$$
g'(u)=\frac{u-\sqrt{2u^2-4}}{\sqrt{2u^2-4}}=\frac{4-u^2}{(u+\sqrt{2u^2-4})\sqrt{2u^2-4}}\leqslant0,
$$
hence $g(u)\leqslant0$ for every $u\geqslant2$. Finally, $f(x)\leqslant1$ for every $x$ and $f(1)=1$ hence $f(x)$ is maximum at $x=1$, and only there.
</s>