23. A very nice exponential inequality (own - from CRUX).

by Virgil Nicula, Apr 22, 2010, 9:50 PM

Virgil Nicula wrote:
Let $a$ and $b$ be two positive real numbers for which $(a-1)(b-1)>0$ . Prove that $\boxed {\ a^b + b^a > 1 + ab + (1-a)(1-b)\cdot\min \{1,ab\}\ }$ .

Lemma (Bernoulli's inequality).

$\blacksquare \ 1^{\circ}.\ 0<a<1\Longrightarrow \left( \forall \right) x>-1,\ (1+x)^a\le 1+ax$ .

$\blacksquare \ 2^{\circ}.\ 1\le a\Longrightarrow \left( \forall \right) x>-1,\ (1+x)^a\ge 1+ax$ .

Proof of the proposed problem.

The first case: $0<a,b\le 1\Longrightarrow \min \{ 1,ab\}=ab$ . Thus, $\left(\frac 1a\right)^b=\left[ 1+\left( \frac 1a -1\right)\right]^b\le \frac{a+b-ab}{a}\Longrightarrow a^b\ge \frac{a}{a+b-ab}$ . Analogously, $b^a\ge \frac{b}{a+b-ab}$.

Thus, $a^b+b^a\ge \frac{a+b}{a+b-ab}=1+ab+\frac{ab(1-a)(1-b)}{a+b-ab}$ . Since $a+b-ab=a+b(1-a)>0,\ \frac{ab}{a+b-ab}\ge ab\Longleftrightarrow (1-a)(1-b)\ge0$ , true.

Thus, $\frac{ab(1-a)(1-b)}{a+b-ab}\ge ab(1-a)(1-b)$ . Therefore, $a^b+b^a\ge 1+ab+ab(1-a)(1-b)=1+ab+(1-a)(1-b)\cdot \min \{1,ab\}$ .

The second case: $1\le a,\ 1\le b \Longrightarrow \min \{1,ab\}=1$ . $a^b\ge 1+b(a-1)$, i.e. $a^b\ge 1+ab-b$. Analogously, $b^a\ge 1+ab-a$ .

Thus, $a^b+b^a\ge 1+ab+(1-a)(1-b)=1+ab+(1-a)(1-b)\cdot \min \{1,ab\}$ .
This post has been edited 8 times. Last edited by Virgil Nicula, Nov 23, 2015, 6:38 AM

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