# How prove $-\sqrt{2}\log(\cos x)\leq\sqrt{x\tan x-\sin^{2}x}$?

How prove that $-\sqrt{2}\log(\cos x)\leq\sqrt{x\tan x-\sin^{2}x}$ for all $x\in\left [ 0,\frac{\pi}{2}\right)$?

• I do not know how to even begin – piteer Aug 7 '14 at 18:04
• Use the derivative – i. m. soloveichik Aug 7 '14 at 21:31

Take $f(x)=-\sqrt{2}\log\cos x$ and $g(x)=\sqrt{x \tan x-\sin^2 x}$. Since $f(0)=g(0)=0$, it is sufficient to show that $f'-g'\leq 0$ over $(0,\pi/2)$. This ultimately boils down to proving: $$4x-\sin(4x)-4\sqrt{2}\sin(2x)\sqrt{x\tan x-\sin^2 x}\geq 0.\tag{1}$$ or the equivalent: $$(4x-4\sin(2x)+\sin(4x))^2\geq 0,\tag{2}$$ that is trivial.