Any idea?

We will repeatedly use the fact that $\int_{0}^{1}(h(x)+h^{-1}(x))~\text{d}x=1$ for any continuous and bijective function $h:[0,1]\to [0,1]$ with $h(0)=0.$ We first make a change of variables: $$\int_{0}^{1}x^{\alpha}(f(x)+f^{-1}(x))~\text{d}x=\frac{1}{\alpha+1}\int_{0}^{1}(f(x)+f^{-1}(x))~\text{d}x^{\alpha+1}=\frac{1}{\alpha+1}\int_{0}^{1}(f(x^{\frac{1}{\alpha+1}})+f^{-1}(x^{\frac{1}{\alpha+1}}))~\text{d}x$$ Let $g:[0,1]\to [0,1]$ be defined by $g(x)=f(x^{\frac{1}{\alpha+1}})$ - this function is continuous, bijective, and $g^{-1}(x)=(f^{-1}(x))^{\alpha+1}$, so $\int_{0}^{1}g(x)~\text{d}x+\int_{0}^{1}g^{-1}(x)~\text{d}x=1$ translates as $$\int_{0}^{1}f(x^{\frac{1}{\alpha+1}})~\text{d}x+\int_{0}^{1}(f^{-1}(x))^{\alpha+1}~\text{d}x=1.$$Similarly, one has $\int_{0}^{1}f^{-1}(x^{\frac{1}{\alpha+1}})~\text{d}x+\int_{0}^{1}f(x)^{\alpha+1}~\text{d}x=1.$ Overall, by AM-GM plus a reversal of the change of variables in the end, we obtain that \begin{align*} \frac{4(\alpha+1)}{\alpha+2} &=\int_{0}^{1}(f(x^{\frac{1}{\alpha+1}})+f^{-1}(x^{\frac{1}{\alpha+1}}))~\text{d}x+\int_{0}^{1}(f(x)^{\alpha+1}+\alpha\cdot x^{\alpha+1})~\text{d}x+\int_{0}^{1}((f^{-1}(x))^{\alpha+1}+\alpha\cdot x^{\alpha+1})~\text{d}x \\ &\ge\int_{0}^{1}(f(x^{\frac{1}{\alpha+1}})+f^{-1}(x^{\frac{1}{\alpha+1}}))~\text{d}x+(\alpha+1)\int_{0}^{1}x^{\alpha}(f(x)+f^{-1}(x))~\text{d}x \\ &= 2(\alpha+1)\int_{0}^{1}x^{\alpha}(f(x)+f^{-1}(x))~\text{d}x \end{align*}which proves the problem.

From wer!!!

Thanks...

see here https://math.stackexchange.com/questions/4115055/how-to-solve-the-following-integral-inequality-alpha2-cdot-int-01-x-alph