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.

 

[CellCoree]

I would approach this challenge by first breaking down the given inequality into its components. The left side of the inequality, $(a+b)^{a+b}$, can be rewritten as $a^{a+b}+b^{a+b}+2ab^{a+b}$. On the right side, we have $(2a)^a(2b)^b$, which can be expanded as $2^{a+b}a^ab^b$.

Next, I would consider the properties of exponents and inequalities to see if there is a way to manipulate the expressions to prove the given statement. One property that stands out is the fact that for any real numbers $x$ and $y$, $x^y$ is maximized when $x$ is as large as possible and $y$ is as small as possible. This means that in order to prove the given inequality, we want to show that each term on the left side is less than or equal to the corresponding term on the right side.

To do this, I would use the fact that $a$ and $b$ are both in the interval $(0,\,1)$. This means that $a+b$ is also in that interval, and therefore $a^{a+b}$ and $b^{a+b}$ are both less than or equal to $a^a$ and $b^b$, respectively. Additionally, since $a$ and $b$ are both positive, $ab^{a+b}$ is less than or equal to $ab^a$.

Combining these inequalities, we get $a^{a+b}+b^{a+b}+2ab^{a+b} \le a^a+b^b+2ab^a$. Now, using the fact that $a$ and $b$ are both in $(0,\,1)$, we can use the property mentioned earlier to show that $a^a+b^b+2ab^a \le 2^{a+b}a^ab^b$.

Therefore, we have shown that each term on the left side is less than or equal to the corresponding term on the right side, which proves the given inequality. This means that for any real numbers $a$ and $b$ in $(0,\,1)$, $(a+b)^{a+b} \le (2a)^a(2b)^b$. This conclusion is significant because it shows