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
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.
It seems that the statement is false even if $f\ge0$ is assumed.
Let $f(x)=x^{10}$. Then $LHS=4(k+1)\cdot\frac1{10k+1}<1<RHS$.
@china math: The analysis problem book doesn't contain this exercise.
