Prove that $a^a+b^b\ge a^b+b^a>1$ if $a,b> 0$.

Question

I succeeded in proving the second part of the inequality by showing that for every real number$x,y ∈ (0, 1)$, we have $x^y≥ \frac x{x + y − xy}.$

By Bernoulli’s inequality we have $x^{1−y}= (1 + x − 1) ^{1−y} ≤ 1 + (x − 1)(1 − y) = x + y − xy,$

$\implies$ $x^y ≥ \frac x{x + y − xy}.$

If $a ≥ 1$ or $b ≥ 1$ then the given inequality clearly holds.

So let $0 <a,b< 1.$ By the previous inequality we have

$a^b + b^a≥ \frac a{a + b− ab} + \frac b{a + b− ab}= \frac {a+b} {a + b− ab} > \frac {a+b}{a+b}= 1.$ But, I found some difficulty on how to prove the first part. Thanks in advance

Without loose of generality, you can suppose 0<a<b<1 . (if a=b, everything holds obviusly). Then take your inequalites and finish — luisfelipe18, Aug 17 '21 at 00:39

Medo · Answer 1 · 2021-08-18T02:22:39.680

4

First, suppose $a>b>1$. Consider the function $f(x):=x^a-x^b$. It suffices to show that $f$ is monotonically increasing on $[1,\infty[$.

But we can write $f(x)=x^b(x^{a-b}-1)$ which is the product of the two strictly positive strictly increasing functions on $]1,\infty[$. So $f$ is monotonically increasing in this case.

Second, Suppose $a>1>b$. Since $a>1$ then $a^{x}$ is strictly increasing and we have $a^a>a^b$. And since $b<1$ then $b^x$ is strictly decreasing and we have that $b^a<b^b$.

Third, Suppose $1>a>b$. Write $f(x):=x^a-x^b=x^a(1-x^{b-a})$ then $f$ is an increasing function on $]0,1[$ (since it is again the product of two strictly positive increasing strictly functions).

The rest follows by symmetry

edited Aug 18 '21 at 02:22

answered Aug 17 '21 at 01:54

Medo

3,588
14
32

I gave a new answer. – Medo Aug 17 '21 at 16:41
In the second case, actually the function $x\mapsto -x^b$ is monotonically decreasing on $[0,+\infty[$. – Antonio Aug 17 '21 at 17:38
Proof of the second case: Since $a>1>b>0$, the function $f(x)=a^x$ is monotonically increasing on $]-\infty,+\infty[$ and the function $g(x)=b^x$ is monotonically decreasing on $]-\infty,+\infty[$. Consequently, $a^a>a^b$ and $b^b>b^a$, hence $a^a+b^b>a^b+b^a$. – Antonio Aug 17 '21 at 17:48
Please, could you correct your proof in the case $a>1>b;$? – Antonio Aug 17 '21 at 17:50
Thanks :) Hey, your persistence pushed me to come up with the right answer. – Medo Aug 18 '21 at 02:22
You are welcome. – Antonio Aug 18 '21 at 04:08

Prove that $a^a+b^b\ge a^b+b^a>1$ if $a,b> 0$.

1 Answers1

Linked