Inequality of integral : $\int_0^{1}((\sqrt{3}f(x))^{2}-2(f(x)^{3})dx≤1$ If $f(x)>0$ be continue function then this inequality true ? 
a) $\int_0^{1}((\sqrt{3}f(x))^{2}-2(f(x)^{3})dx≤1$
b) $(\int_0^{1}\sqrt{3}f(x))^{2}-\int_0^{1}2f(x)^{3}dx≤1$
I was used Cauchy Schwartz inequality but I don't get it 
How I can learning integral inequality if any one have a good book tell me .

If $3-2f(x)<0$, so it's obvious.
But for $3-2f(x)\geq0$ by AM-GM we obtain:
$$\int\limits_0^1(3f(x)^2-2f(x)^3)dx\leq\int\limits_0^1\left(\frac{f(x)+f(x)+3-2f(x)}{3}\right)^3dx=1.$$
Because $$\int\limits_0^1\left(\left(\frac{f(x)+f(x)+3-2f(x)}{3}\right)^3-\left(3f(x)^2-2f(x)^3\right)\right)dx\geq0.$$
