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I'm trying to prove the following inequality $$ \int_{\mathbb{R}} \frac{\lambda}{exp\{-x\} + \lambda} f^2(x) dx \leq log(\frac{1}{\lambda})^{-1} \int_{\mathbb{R}} (1+x^2)^{m} f^{2}(x)dx, $$ where $m$ can be any possible integer and $\lambda\in (0,1)$.

I try to prove the above inequality by showing on $\mathbb{R}$, the following inequality holds \begin{equation} \frac{\lambda}{exp\{-x\} + \lambda} \leq log(\frac{1}{\lambda})^{-1} (1+x^2)^{m}. \end{equation}

To do this, I split the $\mathbb{R}$ in to $S = \{x : log(\frac{1}{\lambda})^{-1} (1+x^2)^{m} \geq 1\}$ and $S^{c}$ be $S$'s Complement in $\mathbb{R}$. Then, the inequality holds over $S$. However, I have difficulty to prove the inequality also holds on $S^{c}$. I have tried to take the regular approach like take the first/second derivative of the function, but it turns out very hard to deal with when the function simultaneously has polynomial and exponential part.

I also use software to check if the inequality holds by picking multiple combination of $\lambda$ and $m$ and all of them seems the inequality holds.

Is there any trick or approach that I can prove the inequality without dealing with the annoying derivative?

Any help is appreciated.

aprita
  • 55

1 Answers1

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Hint with $m$ a positive integer , $a\in(0,1),x\in R^*$:

Using Bernoulli's inequality we have $m\geq 1$:

$$\frac{a}{e^{-x}+a}-\left(\ln\left(\frac{1}{a}\right)\right)^{-1}\left(1+x^{2}\right)^{m}\leq \frac{a}{e^{-x}+a}-\left(\ln\left(\frac{1}{a}\right)\right)^{-1}\left(1+x^{2}m\right)$$

Now using $x+1\leq e^x$ we have and $x\leq 0$:

$$\frac{a}{e^{-x}+a}-\left(\ln\left(\frac{1}{a}\right)\right)^{-1}\left(1+x^{2}m\right)\leq \frac{a}{1-x+a}-\left(\frac{1}{a}-1\right)^{-1}\left(1+x^{2}m\right)$$

For $0\leq x\leq 1$ we have using simple bound :

$$\frac{a}{e^{-x}+a}-\left(\ln\left(\frac{1}{a}\right)\right)^{-1}\left(1+x^{2}m\right)\leq \frac{a}{e^{-x}+a}-\left(\frac{1}{\sqrt{a}}\right)^{-1}\left(1+mx^{2}\right)$$

For $x\geq 1$ we have :

$$\frac{a}{e^{-x}+a}-\left(\ln\left(\frac{1}{a}\right)\right)^{-1}\left(1+x^{2}m\right)\leq \frac{a}{e^{-x}+a}-\left(\frac{1}{a}-1\right)^{-1}\left(1+mx^{2}\right)$$