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