Can the A.M-G.M be manipulated like this?

Question

I was given the following question :

If a + b = 1 , then prove $ a^ab^b + a^bb^a \leq 1 $ , for positive real numbers , a and b.

I proceeded with the A.M-G.M Inequality.

We have :

$$ \frac{(a+a+a....) \, \text{(a times)} + (b+b+b...) \, \text{(b times)}}{a+b} \geq (a^ab^b)^\frac{1}{a+b} $$

This yields $ a^2+b^2 \geq a^a b^b $ as a + b equals one. One can similarly obtain $ 2 a b \geq a^b b^a $ , and adding the two equations should give us the required inequality.

But I quickly realised the error with the proof. To add a to itself a times , when a < 1 , is illogical .

My question is if the proof is right , despite the flaw.

And as a more general question , does

$$ \frac{am + bn}{m+n} \geq (a^mb^n)^\frac{1}{m+n} $$

hold for all positive real numbers a,b,m and n ?

Answer 1

Your proof would work fine for positive integers, but that's not good enough. For the reals, we can prove an even more general theorem using the convexity of the logarithm. This is the weighted AM-GM that dxiv referred to in the comments, and I'll transcribe a proof of it here, for completeness.

Weighted AM-GM: If $\sum\limits_{i=1}^n w_i=w$ and all $w_i, x_i$ are positive reals, then $$\frac{1}{w}\sum_{i=1}^n w_ix_i \geq \sqrt[w]{\prod_{i=1}^nx_i^{w_i}}$$

Proof: The logarithm is concave (since the second derivative is negative), so we have $$\ln\left(\frac{1}{w}\sum_{i=1}^n w_ix_i\right)\geq \frac{1}{w}\sum_{i=1}^nw_i\ln(x_i)=\ln\left(\sqrt[w]{\prod_{i=1}^nx_i^{w_i}}\right)$$ And since the logarithm is strictly increasing, we can get rid of them and preserve the inequality, and here, your result is simply a special case of $n=2$.