4

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 !!!

River Li
  • 49,125
alfred noble
  • 67
  • @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. – alfred noble Oct 25 '16 at 06:15
  • @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 – alfred noble Oct 25 '16 at 06:25
  • @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 – alfred noble Oct 25 '16 at 09:26
  • Where did you come across this interesting inequality? – Thomas Ahle Aug 17 '18 at 16:19

1 Answers1

7

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!) $$ -(e^t (t-2)+t+2)^2\leqslant 0. $$

Fedor Petrov
  • 1,376
  • 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 ? – alfred noble Oct 26 '16 at 06:41
  • 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. – Fedor Petrov Oct 26 '16 at 06:55
  • Yes, I think that this last miraculous identity for the derivative finishes the proof. – Fedor Petrov Oct 26 '16 at 06:58
  • 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:http://math.stackexchange.com/questions/331367/cosh-x-inequality – alfred noble Oct 26 '16 at 16:01
  • I am still not satisfied by your solution. would you please tell me which one is the case in the above inequality when you took the derivatives of both sides of the inequality with respect to $a$? I) the value of $at$ corresponds to the endpoint of the segment or II) a point where the $H'(a) = 0$ – alfred noble Oct 26 '16 at 16:09
  • Endpoints are $a=0,a=1$, this is ok. Rest part of the solution is devoted to the remaining case $H'(a)=0$. – Fedor Petrov Oct 26 '16 at 16:32
  • thanks, what is the value of $a$ ? is it the minimum? how are you sure. It may be any of the function's extrema. Moreover, during your solution you take the derivatives of the sides of inequality with respect to $a$ but later you take another derivative with respect to $t$. Isn't it contradictory ? Finally, what does it mean mathematically when derivatives of both sides of the inequality are equal ? – alfred noble Oct 26 '16 at 17:35
  • I minimize $H(a)$ by $a$ for fixed $t$ and get some function $g(t)$. After that I minimize $g(t)$ by $t$. – Fedor Petrov Oct 26 '16 at 18:18
  • Let us continue this discussion in chat. – alfred noble Oct 27 '16 at 05:27
  • you say that the miraculous inequality at the end of your answer finishes the proof. However how are you sure that all of the operations that you performed are mathematically reversible ? Mathematically you are using reverse approach to prove this inequality therefore all operations must be reversible which means you must be able to get to the main inequality (the posted question) from the final miraculous inequality which you posted online – alfred noble Oct 28 '16 at 13:01
  • Well, you may integrate the miracle from 0 to $t$ to get the inequality in $t$, but is not this unnatural and is it necessary? – Fedor Petrov Oct 28 '16 at 13:14
  • Would you please tell how you are sure that the derivatives of both sides of this inequality, $log(\frac{e^t - 1}{t}) - \frac{t^2}{8} + \frac{t}{e^t - 1 } \le 1 $ is less that 0 ? consider the following function for all $x \gt 0 , f(x) = \frac{1}{x} $ although this function is always greater that 0, the derivative of this function is always less than 0 for all $x \gt 0 $ – alfred noble Oct 30 '16 at 08:18
  • If $f(0)=0$ (wrong for your example) and $f'(t)\leqslant 0$ for $t>0$, then $f(t)\leqslant 0$ for $t\geqslant 0$. – Fedor Petrov Oct 30 '16 at 09:21
  • I did not understand what you mean by your previous comment. As you see for all $x \gt 0 $ we have: $f(x) = \frac{1}{x} \gt 0 $, however for all $x \gt 0 $ we have : $f'(x) = -\frac{1}{x^2} \lt 0 $ which shows that taking the derivatives of both sides of an inequality does NOT necessarily hold the direction of the inequality which means that when you take the derivatives of both sides of $log\frac{e^t - 1}{t} - \frac{t^2}{8} + \frac{t}{e^t-1} \le 1 $ we may NOT necessarily reach to $(log\frac{e^t - 1}{t} - \frac{t^2}{8} + \frac{t}{e^t-1})' \le 0$ ??? !!!! – alfred noble Oct 30 '16 at 11:14
  • 1
    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. – Fedor Petrov Oct 30 '16 at 11:26