prove $f(\sqrt{2})\geq 0$ related $(g(x)f'(x))'+f(x)\geq 0$ Let $g(x)$ has continuous derivative,with $g(x)\geq 1,x\in \mathbf{R}$,if $f(x)\in \mathbf{C^{2}}(-\infty,+\infty)$,and 
$$ f(0)=f'(0)=\int_{-\pi}^{\pi}dy\int_{0}^{\pi}\frac{\cos(nx)-\cos(ny)}{\cos{x}-\cos{y}}dx\qquad n=2k+1,k\in \mathbf{N}^{+} $$
also for and real numbers $x$,
$$(g(x)f'(x))'+f(x)\geq 0 $$
show that:$f(\sqrt{2})\geq 0 $.
My try:
From a know result(American.Mathematical.Monthly.E3145),we get that
$$\int_{0}^{\pi}\frac{\cos{nx}-\cos{ny}}{\cos{x}-\cos{y}}dx=\pi\frac{\sin{ny}}{\sin{y}}\qquad (n=0,1,2\cdots) $$
hence
$$f(0)=f'(0)=\pi\int_{-\pi}^{\pi}\frac{\sin{nx}}{\sin{x}}dx=2\pi^2$$
Then I don't know how to deal with the $$g'(x)f'(x)+g(x)f''(x)+f(x)\geq 0$$.....
 A: First a proof for the case $g(x)=1$:
We have that $f''(x)+f(x)\ge 0$. This can be written as $f''(x)+f(x)=b(x)$ where $b(x)$ is a positive function. Thus:
$$f(x)= k_1 \cos(x)+k_2\sin(x)+\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad$$
$$\quad\quad\quad\cos(x)\int_0^x -b(s)\sin(s)ds+\sin(x)\int_0^x b(s)\cos(s)ds$$
The fact that $f(0)=f'(0)=2\pi^2$ leads to $k_1 = k_2 = 2\pi^2$. So now, by the mean value theorem, since $b(s)$ does not change sign, there exist $a_1,a_2\in[0,\sqrt{2}]$ such that:
$$f(\sqrt{2})=2\pi^2\left(\sin(\sqrt{2})+\cos(\sqrt{2})\right) +\quad\quad\quad\quad\quad\quad\quad\ $$
$$\quad\quad\quad\quad\left(-\cos(\sqrt{2})\sin(a_1)+\sin(\sqrt{2})\cos(a_2)\right) \int_0^\sqrt{2} b(s)ds$$
Thus, we have that:
$$f(\sqrt{2}) \ge 2\pi^2 \ge 0$$
Since the minimal value of $\left(-\cos(\sqrt{2})\sin(a_1)+\sin(\sqrt{2})\cos(a_2)\right)$ is zero.
Now for the case $g(x) \ge 1:$
We define $h'(x) = g(x)f'(x)$. Then again by the mean value theorem, for each $x$, $h(x)= g(a_x) f(x)$, so that $h(x) \ge G f(x)$, where $G$ is the minima of $g(x)$ on the the interval under consideration, which we take to be $[0,\sqrt{2}]$. Thus, we have that:
$$h''(x)+\frac{1}{G}h(x) \ge h''(x)+ f(x) \ge 0 $$
Thus the function $h(x)/\sqrt{G}$ has the same solution as $f(x)$ in the case $g(x)=1$, and using the same logic:
$$h(\sqrt{2})\ge \left(h(0)\cos(\sqrt{2}) + h'(0)\sin(\sqrt{2})\right) \ge$$
$$\left(Gf(0)\cos(\sqrt{2}) + Gf'(0)\sin(\sqrt{2})\right) \ge 2\pi^2 \ge0$$
Note that the actual value of $f(0)=f'(0)$ doesn't make any difference, as long as it's positive.
Q.E.D.
