If $f:[0,1] \rightarrow \mathbb{R},f(0)=0$,is convex and integrable,prove that:$\int_{0}^{1}f(x)dx\ge(2n+1)\int_{0}^{1}(1-x^{\frac{1}{n}})f(x)dx$. If $f:[0,1] \rightarrow \mathbb{R},f(0)=0$,is convex and integrable,prove that:$\int_{0}^{1}f(x)dx\ge(2n+1)\int_{0}^{1}(1-x^{\frac{1}{n}})f(x)dx$.
My progress:
after simplifying I got $2n \int_{0}^{1}f(x) dx \le (2n+1) \int_{0}^{1} f(x)x^{\frac{1}{n}} dx$
Let $\frac{2n}{2n+1} = \lambda , 0 < \lambda <1$ So, I have to show that
$\int_{0}^{1} f(x)x^{\frac{1}{n}}- \lambda f(x) dx \ge 0$ 
 A: Using convexity and the fact that $f(0)=0$, we get that for each $\alpha$ in $[0,1]$, $f\left(\alpha x\right)\leqslant \alpha f(x)$. Therefore, with $\alpha=x^{\frac 1{2n}}$, we have 
$$\int_0^1f(x)x^{1/n}\mathrm dx=\int_0^1f(x)x^{\frac 1{2n}}x^{\frac 1{2n}}\mathrm dx\geqslant\int_0^1f\left(x^{1+\frac 1{2n}}\right)x^{\frac 1{2n}}\mathrm dx.$$
Now use the substitution $u=x^{1+\frac 1{2n}}$ to get the wanted result.
A: This is equivalent to
$$
\int_0^1f(x)\left[x^{1/n}-\frac{2n}{2n+1}\right]\mathrm{d}x\ge0\tag{1}
$$
Let $x_n=\left(\frac{2n}{2n+1}\right)^n$. Then $x_n^{1/n}-\frac{2n}{2n+1}=0$ and $x_n$ decreases toward $e^{-1/2}$.
Note that
$$
\int_0^1x\left[x^{1/n}-\frac{2n}{2n+1}\right]\mathrm{d}x=0\tag{2}
$$
Let $g(x)=f(x)-\frac{x}{x_n}f(x_n)$. Then $g$ is convex, $g(0)=g(x_n)=0$. Therefore,
$$
\begin{align}
g(x)\le0\quad&\text{for }0\le x\le x_n\\
g(x)\ge0\quad&\text{for }x_n\le x\le1
\end{align}\tag{3}
$$
Thus,
$$
g(x)\left[x^{1/n}-\frac{2n}{2n+1}\right]\ge0\tag{4}
$$
Putting together $(2)$, $(3)$, and the integral of $(4)$ yields $(1)$.
