How prove this integral inequality $4(k+1)\int_{0}^{1}(f(x))^kdx\le 1+3k\int_{0}^{1}(f(x))^{k-1}dx$ Question:

let $$f(0)=0,f(1)=1, f''(x)>0,x\in (0,1)$$
  let $k>2$ are real numbers,show that
  $$4(k+1)\int_{0}^{1}(f(x))^kdx\le 1+3k\int_{0}^{1}(f(x))^{k-1}dx$$

This problem is from china Analysis problem book excise by (Min Hui XIE)
,analysis problem bookI think we can use 
$$F(x)=\int_{0}^{x}f(t)dt$$
so
$$F(x)=x\int_{0}^{1}f(ux)du=x\int_{0}^{1}f[ux+(1-u)\cdot 0]du\ge x\int_{0}^{1}[uf(x)dx+(1-u)]du=\dfrac{x}{2}(f(x)+1)$$
then I can't,Thank you
 A: Sign of $\boldsymbol{f}$
If $k\gt2$ is not assumed to be an integer, then we need to assume that $f$ is non-negative for $f(x)^k$ to make sense. Moreover, if $k=4$ and $f(x)=2x^6-x$, then
$$
4(k+1)\int_0^1f(x)^k\,\mathrm{d}x=\frac45\gt\frac{102}{133}=1+3k\int_0^1f(x)^{k-1}\,\mathrm{d}x\tag{1}
$$
Therefore, I think it is safe to assume that $f$ needs to be non-negative.

Variational Argument
To maximize $\int_0^1\left(4(k+1)f(x)^k-3kf(x)^{k-1}\right)\,\mathrm{d}x$ for all $f$ so that $f(0)=0$ and $f(1)=1$ we need
$$
\begin{align}
0
&=\delta\int_0^1\left[4(k+1)f(x)^k-3kf(x)^{k-1}\right]\,\mathrm{d}x\\
&=\int_0^1\left[4(k+1)kf(x)^{k-1}-3k(k-1)f(x)^{k-2}\right]\delta f(x)\,\mathrm{d}x\tag{2}
\end{align}
$$
for every $\delta f$ that fixes $\int_0^1f'(x)\,\mathrm{d}x$; that is,
$$
\begin{align}
0
&=\delta\int_0^1f'(x)\,\mathrm{d}x\\
&=\int_0^1f''(x)\delta f(x)\,\mathrm{d}x\tag{3}
\end{align}
$$
To satisfy these conditions for all $\delta f$, we need a constant $\lambda$ so that
$$
\begin{align}
f''(x)
&=\lambda\left[4(k+1)kf(x)^{k-1}-3k(k-1)f(x)^{k-2}\right]\\
&=\lambda 4k(k+1)f(x)^{k-2}\left[f(x)-\frac34\frac{k-1}{k+1}\right]\tag{4}
\end{align}
$$
Now we can use the condition that $f''(x)\ge0$. Note that if $\lambda\ne0$, then the right side of $(4)$ changes sign as $f(x)-\frac34\frac{k-1}{k+1}$ does, which it must since $f(0)=0$ and $f(1)=1$. Therefore, we must have that $\lambda=0$. This in turn implies that $f''(x)=0$. Thus, we must have that
$$
f(x)=x\tag{5}
$$
which means
$$
\begin{align}
\int_0^1\left(4(k+1)f(x)^k-3kf(x)^{k-1}\right)\,\mathrm{d}x
&\le\int_0^1\left(4(k+1)x^k-3kx^{k-1}\right)\,\mathrm{d}x\\
&=1\tag{6}
\end{align}
$$
which is equivalent to the condition sought.
A: It seems that the statement is false even if $f\ge0$ is assumed. 
Let $f(x)=x^{10}$. Then $LHS=4(k+1)\cdot\frac1{10k+1}<1<RHS$.
A: @china math: The analysis problem book doesn't contain this exercise.
