Let be $f:\left[0,1\right]\rightarrow\left[0,1\right]$ a continuous and bijective function,such that : $f\left(0\right)=0$.Then the following inequality holds: $\left(\alpha+2\right)\cdotp\int_{0}^{1}x^{\alpha}\left(f\left(x\right)+f^{-1}\left(x\right)\right)\leq2,\forall\alpha\geq0 $
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Tags: integral inequalities, college contests
26.04.2021 00:07
Any idea?
26.04.2021 11:16
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.
26.04.2021 16:41
From wer!!!
27.04.2021 09:45
Thanks...
28.04.2021 00:08
see here https://math.stackexchange.com/questions/4115055/how-to-solve-the-following-integral-inequality-alpha2-cdot-int-01-x-alph