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Hello during a problem I come across an another problem :

Let $x>0$ and $f(x)$ a continuous stricly increasing function such as $f(x)-x\geq m$ with $m$ a real number and $$x+\frac{1}{x}\leq f(x)+\left(\frac{1}{x}\right)$$

Prove that we have : $$f(x)f\left(\frac{1}{x}\right)\geq 1 \quad \forall x>0$$

The original problem was the following :

$$\cosh(x)\leq e^{\frac{x^2}{2}}\quad \forall x>0$$

If we put $y=e^x$ this has the form of the beginning with :

$$f(x)=x^{\ln(x)0.5}$$

Many thanks for your contributions.

max8128
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1 Answers1

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You are right, there exists a counterexample. Put $\Bbb R_+=\{x\in\Bbb R: x>0\}$ and define a function $f:\Bbb R_+\to\Bbb R_+$ by putting

$$f(x)= \cases{x/2,\;\;\;\;\;\;\;\;\;\;\; \mbox{if $x\le 1/2$},\\ (7x-3)/2, \mbox{if $1/2\le x\le 1$ },\\ x+1\;\;\;\;\;\;\;\,, \mbox{ if $x\ge 1$.}}$$

Then the function $f$ is continuous and strictly increasing, $f(x)-x\ge -1/4$ and $x+\frac{1}{x}\leq f(x)+f(\frac{1}{x})$ for each $x$, but $f(2)f(1/2)=3/4<1$.

Alex Ravsky
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    Yes it works (+1) but I would like to discuss with you what's happend if we exclude the angular point ? And what's happend if the function is differentiable everywhere ? Thanks for your interest . –  Nov 09 '17 at 20:28
  • @FatsWallers I think these details are not essential and can be easily obtained. The problem arises because $f(x)>0$ can be very small for $x<1$ while assuring all other conditions by sufficiently large $f(x)$ for $x\ge 1$. For instance, if $f(x)\ge 0$ for each $x$ and $f(x)\ge x+1$ for each $x\ge 1$ then already $f(x)+f(1/x)\ge x+1/x$ and $f(x)-x\ge -1$ for each $x$. – Alex Ravsky Nov 09 '17 at 20:40
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    Yes you are right ! A last question :) Can you prove there exists an interval where $f(x)f(\frac{1}{x})\geq 1$.I think it's very difficult ^^.Thanks for your sagacity . –  Nov 09 '17 at 21:28
  • @FatsWallers You are right again, because this is impossible. Put $f(x)= x-1/x+1$, if $x\ge 1$ and $f(x)=2x-1$, if $x\le 1$. Then $x+\frac{1}{x}=f(x)+f(\frac{1}{x})$, but for $x\ge 1$ we have $f(x)f(1/x)\ge 1$ iff $(1-x)(x^2+x-2)\ge 0$, that is iff $x=1$. – Alex Ravsky Nov 10 '17 at 05:32
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    Good, I have an other problem wich is difficult .Have you some idea on this ? https://math.stackexchange.com/questions/2512249/inequality-with-power-and-logarithm .Have a good day. –  Nov 10 '17 at 08:59
  • @FatsWallers Done. PS. It was not difficult. – Alex Ravsky Nov 10 '17 at 14:57