Today,when I use wolf found this following inequality
let $x>-1$, show that $$\dfrac{e^x}{x+1}>\dfrac{\cos{x}}{\sin{x}+\sqrt{2}}$$
I found this 
I want $$\Longleftrightarrow (\sin{x}+\sqrt{2})e^x-(x+1)\cos{x}>0$$ let $$f(x)=(\sin{x}+\sqrt{2})e^x-(x+1)\cos{x}\Longrightarrow f'(x)=\sqrt{2}e^x+(x+e^x+1)\sin{x}+(e^x-1)\cos{x}$$ then I calculus $f''(x)$ and found ugly,so maybe this inequality have other methods?
I think it just follows from: $$\left|\frac{\cos x}{\sqrt{2}+\sin x}\right|\leq 1\tag{1}$$ and: $$\frac{e^x}{x+1}\geq 1.\tag{2}$$ $(1)$ is equivalent to: $$ \cos^2 x\leq 2+\sin^2 x+2\sqrt{2}\sin x \tag{3}$$ or to: $$ 1+2\sin^2 x+2\sqrt{2}\sin x \geq 0 \tag{4}$$ that is trivial since the discriminant of $2y^2+2\sqrt{2}y+1$ is zero. On the other hand, $$ e^x \geq 1+x \tag{5} $$ is a trivial inequality, too.
We know $\sqrt{2}+\sin x > 0$ since $|\sin x| \le 1$ and $\sqrt{2} > 1$.
$\displaystyle \frac{e^{x}}{x+1} \ge 1 \ge \frac{\cos x}{\sqrt{2}+\sin x}$, the later inequality is true since,
$1 \ge \sin (\frac{\pi}{4} - x) = \dfrac{\cos x - \sin x}{\sqrt{2}}$
