<s>[INST] Proving $\sqrt{2}x-\sqrt{x^{2}+1} \geq \frac{\sqrt{2}}{2}\ln{(x)}$ How can I prove that
$$
x\sqrt{2}-\sqrt{x^{2}+1} \geq \frac{\sqrt{2}}{2}\ln{(x)}
$$
It's a derivation-based process if I remember correctly, however I was unable to prove it correctly.
 [/INST] $$f(x):=x\sqrt{2}-\sqrt{x^2+1}-\frac{\sqrt{2}}{2}\ln x$$
Which is continuously differentiable on $(0,\infty)$
$$f'(x)=-\frac{x}{\sqrt{x^2+1}}-\frac{1}{\sqrt{2} x}+\sqrt{2}=\frac{-2 x^2+2 \sqrt{2} \sqrt{x^2+1} x-\sqrt{2} \sqrt{x^2+1}}{2 x \sqrt{x^2+1}}$$
We want to solve 
$$
-2 x^2+2 \sqrt{2} \sqrt{x^2+1} x-\sqrt{2} \sqrt{x^2+1}=0$$
$$
\frac{2x^2}{2x-1}=\sqrt{2}\sqrt{x^2+1}\Rightarrow -2 + 8 x - 10 x^2 + 8 x^3 - 4 x^4=0
$$
Or:
$$
(x-1) \left(2 x^3-2 x^2+3 x-1\right)=0
$$
The second term has two complex roots, as its discriminant is negative, and a root between $0$ and $\frac{1}{2}$(by Bolzano-Weierstrass), but from 
$$
\frac{2x^2}{2x-1}=\sqrt{2}\sqrt{x^2+1}
$$
we can see that it must be a false root. We have to check the second derivative:
$$f''(x)=\frac{x^2}{\left(x^2+1\right)^{3/2}}-\frac{1}{\sqrt{x^2+1}}+\frac{1}{\sqrt{2} x^2}$$
$$\left.f''(x)\right|_{x=0}=\frac{1}{2\sqrt{2}}>0$$
Thus it is a minimum, and the only possible one.
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