Inequality Challenge V: Prove $(a+b)^{a+b} \le (2a)^a(2b)^b$

[anemone]

Prove that for any real numbers $a$ and $b$ in $(0,\,1)$, that $(a+b)^{a+b}\le (2a)^a(2b)^b$.

 

[Opalg]

You probably want an algebraic proof of this, but as an analyst I naturally think in terms of an analytic proof.
[sp]Dividing both sides by $2^{a+b}$, we need to show that $\Bigl(\dfrac{a+b}2\Bigr)^{a+b} \leqslant a^ab^b$. Then taking the square root of both sides, we need to show that $\bigl(\frac12(a+b)\bigr)^{(a+b)/2} \leqslant \sqrt{a^ab^b}.$ Taking logs of both sides, we need to show that $\bigl(\frac12(a+b)\bigr) \ln\bigl(\frac12(a+b)\bigr) \leqslant \frac12(a\ln a + b\ln b).$ But that is an immediate consequence of the fact the function $f(x) = x\ln x$ is concave, so that $f\bigl(\frac12(a+b)\bigr) \leqslant \frac12\bigl(f(a) + f(b)\bigr).$

To check that $f$ is concave, notice that $f'(x) = \ln x + 1$, $f''(x) = 1/x >0$ for all $x>0$.

This proof shows that the result holds for all positive $a$ and $b$, not just those in the interval $(0,1)$, and that equality holds only when $a=b$.[/sp]

 

[anemone]

Thanks, Opalg for your neat solution in tackling this challenge problem. I'll post the solution (half-algebraic half-analytic) sometime later!:)

 

[anemone]

Spoiler

If we write $a=t(a+b)$ and $b=(1-t)(a+b)$ so that $0<t<1$ and taking both sides of the inequality to the power $\dfrac{1}{a+b}$ and dividing by $a+b$, the inequality is equivalent to

$1\le(2t)^t(2(1-t))^{1-t}$

$\log \dfrac{1}{2}\le t\log t+(1-t)\log(1-t)$

Let $f(t)$ denotes the function on the right then we have $f'(t)=\log t-log(1-t)$, which is negative if $0<t<\dfrac{1}{2}$, equals to 0 at $t=\dfrac{1}{2}$, and positive if $\dfrac{1}{2}<t<1$.

Thus $f(t)$ is minimal at $t=\dfrac{1}{2}$, and since $f\left( \dfrac{1}{2} \right)=\left( \dfrac{1}{2} \right) \log \left( \dfrac{1}{2} \right)+\left( 1-\dfrac{1}{2} \right)\log \left( 1-\dfrac{1}{2} \right)=\log \left(\dfrac{1}{2} \right)$ and hence we proved for the desired inequality.