<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] Let $f(x)=x\sqrt{2}-\sqrt{x^{2}+1} -\dfrac{\sqrt{2}}{2}\ln{(x)}$. Then, $$f'(x)=\sqrt{2}-\frac{x}{\sqrt{x^2+1}}-\frac{1}{x\sqrt{2}}=\frac{2x\sqrt{x^2+1}-x^2\sqrt{2}-\sqrt{x^2+1}}{x\sqrt{2(x^2+1)}}.$$
For $x>0$, the denominator is positive, so we just need to check the sign of the numerator: $$2x\sqrt{x^2+1}-x^2\sqrt{2}-\sqrt{x^2+1}\geq2\sqrt{2}-\sqrt{2}-\sqrt{2}=0  \;\;\text{ for } \; x\geq1.$$
Thus, $f(x)$ is an increasing function on $[1,\infty)$. 
Since $f(1)=0$, it follows that $f(x) \geq 0$ for all $x \in [1,\infty)$. Hence, $$x\sqrt{2}-\sqrt{x^{2}+1} -\dfrac{\sqrt{2}}{2}\ln{(x)} \geq0 \implies x\sqrt{2}-\sqrt{x^{2}+1} \geq \frac{\sqrt{2}}{2}\ln{(x)}.$$
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