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.
 A: 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 $1$).
A: 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, $$x\sqrt{2}-\sqrt{x^{2}+1} -\dfrac{\sqrt{2}}{2}\ln{(x)} \geq0 \implies x\sqrt{2}-\sqrt{x^{2}+1} \geq \frac{\sqrt{2}}{2}\ln{(x)}.$$
A: $$f(x):=x\sqrt{2}-\sqrt{x^2+1}-\frac{\sqrt{2}}{2}\ln x$$
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.
