How prove this integral inequality $2\int_{-1}^{1}f(x)g(x)dx\ge\int_{-1}^{1}f(x)dx\int_{-1}^{1}g(x)dx$ Let $f:[-1,1]\longrightarrow \mathbb{R}$ be increasing on $[0,1]$ and even, i.e. $f(x)=f(-x)$ $\forall x\in [-1,1]$.
Let $g:[-1,1]\longrightarrow \mathbb{R}$ be convex, i.e. $g(tx+(1-t)y)\le tg(x)+(1-t)g(y)$
for every $x,y\in [-1,1]$ and $t\in (0,1)$,  
Show that

$$2\int_{-1}^{1}f(x)g(x)dx\ge\int_{-1}^{1}f(x)dx\int_{-1}^{1}g(x)dx$$

My try: I think this can use Cauchy-Schwarz inequality

$$\int_{a}^{b}f^2(x)dx\int_{a}^{b}g^2(x)dx\ge\left(\int_{a}^{b} f(x)g(x)dx\right)^2$$

But I can't, and this problem is from China competition yesterday.

I found this (written in Chinese):

