How to show that $a^ab^b+a^bb^a\leq 1$ for $a,b\in(0,1)$ with $a+b=1$. [duplicate]

A little while ago I came across this question on Yahoo! Answers and it had no replies. I have spent a while trying to figure this out as well, but I couldn't find an algebraic proof.
The assumptions are $a+b=1$ and $a,b\in(0,1)$. We want to show that $a^ab^b+a^bb^a\leq 1$. Or equivalently that $a^a(1-a)^{1-a}+a^{1-a}(1-a)^a\leq 1\;\forall a\in(0,1)$.

I tried see if $a$ and $b$ are an upper bound for $a^ab^b$ and $a^bb^a$ respectively, for their sum would be $1$ by assumption. However, I was unable to do so. I managed to show the following, assuming without loss of generality that $b\leq a$ (i.e. $a\geq\frac{1}{2}$): $$\begin{aligned} a^ab^b&=a^{1-b}b^b\\ &=a\left(\frac{b}{a}\right)^b\\ &\leq a. \end{aligned}$$ However, applying this same trick to $a^bb^a$ does not yield the necessary inequality: $$\begin{aligned} a^bb^a&=a^bb^{1-b}\\ &=b\left(\frac{a}{b}\right)^b\\ &\geq b. \end{aligned}$$ Which means that the upperbound $a^bb^b\leq a$ is too weak.
By similar means one can find the following bounds: $$b\leq a^ab^b\leq a,$$ $$b\leq a^bb^a\leq a.$$

Now define $f(a):=a^a(1-a)^{1-a}=a^ab^b$. One can show that the following holds for the derivative of $f$: $$f'(a)=f(a)\log\left(\frac{a}{1-a}\right).$$ Since $f(a)=\exp(a\log(a)+(1-a)\log(1-a))>0$, we have that $f'(a)=0\iff a=\frac{1}{2}$. Since $f'(a)>0$ for all $a>\frac{1}{2}$ we have that $f$ has a minimum at $a=\frac{1}{2}$. Also note that $\lim_{a\to 1}f(a)=1$, hence by symmetry $\frac{1}{2}\leq f(a)\leq 1$ for all $a\in(0,1)$.

We'll define $g(a):=a^{1-a}(1-a)^a=a^bb^a$. One finds that $$g'(a)=g(a)\left(\frac{1-2a}{a(1-a)}+\log\left(\frac{1-a}{a}\right)\right).$$ Now $g'(a)=0\iff\log\left(\frac{1-a}{a}\right)=\frac{2a-1}{a(1-a)}$. It's clear that $a=\frac{1}{2}$ is a solution. Note that $\frac{d}{dx}\log\left(\frac{1-a}{a}\right)=\frac{1}{a(a-1)}<0$. Since $\frac{2a-1}{a(1-a)}$ is increasing, $a=\frac{1}{2}$ is the only solution. As $\lim_{a\to 1}g(a)=0$, we have that this is a maximum. Hence $0\leq g(a)\leq\frac{1}{2}$ for all $a\in(0,1)$.

Combining these results with what we already had yields the following inequalities: $$\frac{1}{2}\leq a^ab^b\leq a,$$ $$b\leq a^bb^a\leq\frac{1}{2}.$$ This gives us $$b+\frac{1}{2}\leq a^ab^b+a^bb^a\leq a+\frac{1}{2}.$$ Which is still not all too helpful... So then I got another idea. I construct two parabolas $\phi,\psi$ both with their top at $\left(\frac{1}{2},\frac{1}{2}\right)$. One of them, $\phi$, through $(0,0)$ and $(1,0)$ and $\psi$ through $(0,1)$ and $(1,1)$. Then I'd show that $a^a(1-a)^{1-a}\leq\psi(a)$ and $a^{(1-a)}(1-a)^a\leq\phi$. Then, since $\phi(a)+\psi(a)=1\;\forall a$ the statement is proven. However, this created some difficulty that I am now stuck on. I'll first define $\phi$ and $\psi$: $$\phi:(0,1)\to\left(0,\frac{1}{2}\right]$$ $$x\mapsto-2x^2+2x$$ Then $\psi(x)=1-\phi(x)$. We have $$\begin{aligned} \phi(a)&\geq a^{1-a}(1-a)^a\\ \iff-2a^2+a&\geq a^{1-a}(1-a)^a\\ \iff2a&\geq\left(\frac{a}{1-a}\right)^{1-a}. \end{aligned}$$ I cannot figure out how to prove this inequality though; let alone the other one. Hence now I am stuck!

I'm hoping someone can help figure this out. I realize parts of what I've done may be superfluous, but this is just everything that I have tried put together.

Solution 1:

We have $$1 = a + b = a^{a+b} + b^{a+b} = a^{a}a^{b} + b^{a}b^{b}$$ So we have, $$1 - a^{a}b^{b} - a^{b}b^{a} = a^{a}a^{b} + b^{a}b^{b} - a^{a}b^{b} - a^{b}b^{a} = (a^{a} - b^{a})(a^{b} - b^{b})$$ Now if $a\leq b$, then $a^{a}\leq b^{a}$ and $a^{b}\leq b^{b}$. If $a\geq b$, then $a^{a}\geq b^{a}$ and $a^{b}\geq b^{b}$.

So, in both cases R.H.S is positive. Hence, $a^{a}b^{b} + a^{b}b^{a}\leq 1$ with equality holding iff $a = b = \frac {1}{2}$.

Solution 2:

Your last inequality can be also solved if you go ahead.

it is equal to $f(x)=x^x(1-x)^{1-x} \ge \dfrac{1}{2}$ or we try to find $f_{min}$.

Let $g(x)=ln(f(x))=xlnx+(1-x)ln(1-x)$

$g'(x)=ln\dfrac{x}{1-x}=0 \implies \dfrac{x}{1-x}=1 \implies x=\dfrac{1}{2}$

it is easy to verify $g(\dfrac{1}{2})$ is min $\implies f(\dfrac{1}{2})$ is min also.

Edit: I repeat all the process which op did in his post. I add a graphic for fun.

red one is $f(x）$, green one is $h(x)=x^{1-x}(1-x)^x$, blue one is $k(x)=f(x)+h(x)$

so this function also have min bound which $ >0.845$ when $x=0.118$ or $x=1-0.118$