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Let $0<x\leq 1$ then :

$$f(x)=\left(\frac{1}{x}\right)^{\frac{2}{x^2+1}}+\frac{1}{x}-\frac{1}{x^2}$$

The claim is :

$$f''(x)\geq 0$$

I cannot show it but I can say what is wrong or true as way :

First we can differentiate twice and try to find some inequalities to simplify the derivative :

1)Replace $\left(\frac{1}{x}\right)^{\frac{2}{x^2+1}}$ by $\frac{1}{x}$ doesn't works .

2)Replace the exponent $\frac{2}{x^2+1}$ by $2-x$ works for $0.8<x\leq 1$ wich is the hardest part.So using binomial theorem for real exponent we can perhaps delete the exponent

3)We can use the Bernoulli's inequality: $$\left(\frac{1}{x}\right)^{\frac{2}{x^{2}+1}}\geq \left(\frac{1}{x}-1\right)\left(\frac{2}{x^{2}+1}\right)+1$$ And replaces it in the second derivative with the assumption $0.75\leq x\leq 1$

If you have some others way to tackle it like Popoviciu's inequality I accept too !

Question :

How to show the claim ?

Miss and Mister cassoulet char
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1 Answers1

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Just some preliminary results.

We have $$f''(x)=\frac{2}{x^4 \left(x^2+1\right)^4}\, g(x)$$ $$g(x)=(x-3) \left(x^2+1\right)^4+x^{\frac{2 x^2}{x^2+1}} \Bigg[8 x^4 \log ^2(x)+\left(x^2+1\right)^2 \Big[3+x^2 (5-6 \log (x))\Big]\Bigg]$$ At the end points, we have $$g(0)=0^+ \qquad \text{and} \qquad g'(0)=1^-\qquad \text{and} \qquad g''(0)=-2$$ $$g(1)=0^+ \qquad \text{and} \qquad g'(1)=0^-\qquad \text{and} \qquad g''(0)=16$$

So, in the range, there is a maximum value of $g(x)$. Continuing with the big term in brackets, it is an increasing funtion of $x$.

Expanding $g(x)$ as a series around $x=\frac 12$ shows that the maximum value is close to this point.

What would remain to prove is that there is only one solution for $g'(x)=0$.

Claude Leibovici
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