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
Simply consider that $\sqrt{x}$ is a concave function, hence, for any $a,b\geq 0$, $\sqrt{a}+\sqrt{b}\leq \sqrt{2(a+b)}$. By taking $a=x$ and $b=\frac{1+x^2}{2}$ you get $f(x)\leq 1$ with equality only when $x=\frac{1+x^2}{2}$, i.e. $x=1$, as wanted.
Find f''(x) 

if  f''(x_0) is -ve than x_0 is point of maximum 

else f''(x_0) is  +ve than x_0 is point of minimum