I am trying to solve the following question, but I did not reach to any answer, I would be sol glad if anyone could help me on that.

If $a, b$ are positive numbers such that $a + b = 1$, prove that for all real numbers $x$, $$ae^{\frac{x}{a}} + be^{-\frac{x}{b}}\leq e^{\frac{x^2}{8a^2b^2}}.$$

Thank you everyone !!!

  • $\begingroup$ @Michael would you please devote some time to prove this inequality. I would really appreciate it if you mind helping me or giving me some hints on how to deal with inequalities like the one that I posted above. $\endgroup$ Commented Oct 25, 2016 at 6:15
  • $\begingroup$ @Michael, if we consider the case where $a = b = \frac{1}{2}$ then we have: $cosh(2x) \le e^{2x^2}$. I would work on proving the latter inequality, this method may help to prove the main inequality $\endgroup$ Commented Oct 25, 2016 at 6:25
  • $\begingroup$ @πr8, I read your answer which proves this above inequality when $a = b = \frac{1}{2}$, now I would deeply appreciate it if you could help prove the above inequality in general form $\endgroup$ Commented Oct 25, 2016 at 9:26
  • $\begingroup$ Where did you come across this interesting inequality? $\endgroup$ Commented Aug 17, 2018 at 16:19

1 Answer 1


It looks to be neat but true. Denote $x=abt$, we rewrite inequality as $ae^{bt}+be^{-at}\leqslant e^{t^2/8}$. We may suppose $t>0$, else replace $t$ to $-t$ and $b$ to $a$. Multiply by $e^{at}$ and rewrite as $ae^t+1-a\leqslant e^{at+t^2/8}$. Fix $t$ and vary $a\in [0,1]$. We should consider the minimal value of $H(a)=e^{at+t^2/8}-ae^t-1+a$ and prove that it is non-negative. This minimal value is attained either for $a=0$, or $a=1$, or for such $a$ that $H'(a)=0$. Obviously $H(0)\geqslant 0$, $H(1)\geqslant 0$, thus it remains to consider such $a$ that $H'(a)=0$. That is, $e^t-1=te^{at+t^2/8}$, $at=\log((e^t-1)/t)-t^2/8$. Now the inequality may be rewritten (multiply it by $t$ and substitute the values for $at$ and for $te^{at+t^2/8}$) as $$(e^t-1)\left(\log\frac{e^t-1}t-\frac{t^2}8\right)+t\leqslant e^t-1,$$ Divide by $e^t-1$ and note that for $t=0$ the equality takes place, so it suffices to prove that $$\left(\log\frac{e^t-1}t-\frac{t^2}8+\frac{t}{e^t-1}\right)'\leqslant 0,\,\forall t\geqslant 0.$$ Taking derivative and multiplying by $4t(e^t-1)^2$ we get (miracle!) $$ -(e^t (t-2)+t+2)^2\leqslant 0. $$

  • $\begingroup$ thanks a lot. But I have two questions about your answer. I) In the middle of your answer you took the derivatives of both sides of the inequality with respect to $a$ and you calculated the value of $a$ for which the derivations of both sides of the inequality is the same. I want to know why did you do that ? II) Are you saying that since you reached to an ever true inequality ($-(e^t(t-2) + t + 2)^2 \le 0$) the main inequality that I posted above is also true ? $\endgroup$ Commented Oct 26, 2016 at 6:41
  • $\begingroup$ If we need to prove that $H(a)\geqslant 0$ for some function $H$ (here $H=$RHS-LHS of the inequality) and $a\in [0,1]$, we look at a point $a\in [0,1]$, for which $H$ takes minimal value. It is either an endpoint of the segment, or a point where derivative $H'(a)$ equals 0. $\endgroup$ Commented Oct 26, 2016 at 6:55
  • $\begingroup$ Yes, I think that this last miraculous identity for the derivative finishes the proof. $\endgroup$ Commented Oct 26, 2016 at 6:58
  • $\begingroup$ This is the link to the prove of the above inequality in special case when $a = b = \frac{1}{2}$ I think it might be useful to prove the above inequality in the general form. LINK:math.stackexchange.com/questions/331367/cosh-x-inequality $\endgroup$ Commented Oct 26, 2016 at 16:01
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    $\begingroup$ Additional requirement is $f(0)=0$. For $f(x)=1/x$ it is not true, but for $-1+\log\frac{e^x-1}x-x^2/8+\frac{x}{e^x-1}$ it is true. $\endgroup$ Commented Oct 30, 2016 at 11:26

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