TITLE: How prove this integral inequality $6\left(\int_{0}^{1}f(x)dx\right)^2\le 1+ 8\int_{0}^{1}f^3(x)dx$
QUESTION [7 upvotes]: Let $f$ be a positive-valued,concave function on $[0,1]$,Prove that
  $$6\left(\int_{0}^{1}f(x)dx\right)^2\le 1+ 8\int_{0}^{1}f^3(x)dx$$

Let $$A=\int_{0}^{1}f^3(x)dx,B=\left(\int_{0}^{1}f(x)dx\right)^2$$
$$\Longleftrightarrow 6B\le 1+8A$$
let $$F(x)=\int_{0}^{x}f(t)dt$$
$$F(x)=x\int_{0}^{1}f[ux+(1-u)\cdot 0]du\ge x\int_{0}^{1}[uf(x)+(1-u)du=\dfrac{xf(x)}{2}+\dfrac{x}{2}$$
Maybe this problem folowing  can use Cauchy-Schwarz inequality to solve it,Thank you

REPLY [6 votes]: Let $c=\int_0^1 f(x)\,dx$ and $g=f/c$, so $\int_0^1 g(x)\,dx=1$. Then by Holder's inequality,
$$
1\le \left(\int_0^1 g(x)^3\,dx\right)^{1/3}\left(\int_0^1 1^{3/2}\,dx\right)^{2/3} .
$$
Therefore $\int_0^1 f(x)^3\,dx=c^3\int_0^1g(x)^3\,dx\ge c^3$, and 
$$
8\int_0^1 f(x)^3\,dx + 1-6\left(\int_0^1 f(x)\,dx\right)^2\ge 8c^3+1-6c^2 =: h(c).
$$
For $c>0$ the right-hand side is minimized when $0=h'(c)=24c^2-12c$, meaning $c=1/2$ (noting $h'(c)<0$ for $c<1/2$ and $h'(c)>0$ for $c>1/2$). Thus $$h(1/2)=8(1/2)^3+1-6(1/2)^2=1/2\le h(c)$$ for all $c>0$.
Actually, then it follows
$$
6\left(\int_0^1 f(x)\,dx\right)^2 \le \frac12 + 8\int_0^1f(x)^3\,dx.
$$
Concavity of $f$ is not needed.