Show that $f(x)=2x+\cot^{–1}x+\log(\sqrt{1+x^2}−x)$ is increasing in $\mathscr{R}$.
My Attempt $$ f'(x)=2-\frac{1}{1+x^2}+\frac{\frac{x}{\sqrt{1+x^2}}-1}{\sqrt{1+x^2}−x}\\ =2-\frac{1}{1+x^2}+\frac{\frac{x-\sqrt{1+x^2}}{\sqrt{1+x^2}}}{\sqrt{1+x^2}−x}\\ =2-\frac{1}{1+x^2}-\frac{1}{\sqrt{1+x^2}}\\ =\frac{2(1+x^2)-1-\sqrt{1+x^2}}{1+x^2}\\ =\frac{1+2x^2-\sqrt{1+x^2}}{1+x^2} $$ How do I show that $1+2x^2\geq\sqrt{1+x^2}\implies f'(x)\geq 0$ and thus the given function is increasing in $\mathscr{R}$ ?
HINT:
For $x\neq0$, $$1+2x^2>\sqrt{1+x^2}\impliedby1+4x^2+4x^4>1+x^2\impliedby 4x^4+3x^2>0$$