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.

Let's get rid of $\sqrt{2}$ by rewriting the inequality as
$$
f(x)=2x-\sqrt{2(x^2+1)}-\ln x\ge0
$$
We have $\lim_{x\to0}f(x)=\infty$. For computing the limit at $\infty$, we do the substitution $t=1/x$, so the limit becomes
$$
\lim_{t\to0^+}\frac{2}{t}-\frac{\sqrt{2(t^2+1)}}{t}+\ln t=
\lim_{t\to0^+}\frac{2-\sqrt{2(t^2+1)}+t\ln t}{t}=\infty
$$
Thus we know that $f$ has at least a point of minimum.
Compute the derivative
$$
f'(x)=2-\frac{2x}{\sqrt{2(x^2+1)}}-\frac{1}{x}
=2-\frac{x\sqrt{2}}{\sqrt{x^2+1}}-\frac{1}{x}
$$
Let's go on:
$$
f''(x)=-\sqrt{2}\,\frac{\sqrt{x^2+1}-\dfrac{x^2}{\sqrt{x^2+1}}}{x^2+1}
+\frac{1}{x^2}
=-\sqrt{2}\,\frac{1}{(x^2+1)\sqrt{x^2+1}}+\frac{1}{x^2}
$$
and we want to evaluate the sign of
$$
(x^2+1)\sqrt{x^2+1}-x^2\sqrt{2}
$$
that is, where $(x^2+1)\sqrt{x^2+1}>x^2\sqrt{2}$. We can square getting
$$
x^6+3x^4+3x^2+1>2x^4
$$
that is
$$
x^6+x^4+3x^2+1>0
$$
which is of course true.
Therefore $f''(x)>0$ for all $x>0$ and so $f'(x)$ is increasing. Since $f'(1)=0$, $f'$ vanishes only at $1$, which is thus the unique minimum point for $f$.
Since $f(1)=0$, we see that $f(x)\ge0$ for all $x>0$ (equality only at $1$).
