Is there a good reason why a^{2b} + b^{2a} <= 1 when a+b=1? - MathOverflow

Is there a good reason why a^{2b} + b^{2a} <= 1 when a+b=1?

2010-03-05T16:18:14Z

The following problem is not from me, yet I find it a big challenge to give a nice (in contrast to 'heavy computation') proof. The motivation for me to post it lies in its concise content.

If $a$ and $b$ are nonnegative real numbers such that $a+b=1$, show that $a^{2b} + b^{2a}\le 1$.

Answer by Mariano Suárez-Alvarez for Is there a good reason why a^{2b} + b^{2a} <= 1 when a+b=1?

2010-03-05T16:27:06Z

Silly non-answer removed.

Answer by Markus for Is there a good reason why a^{2b} + b^{2a} <= 1 when a+b=1?

2010-03-05T17:32:22Z

EDIT: wrong proof attempt

Answer by Gjergji Zaimi for Is there a good reason why a^{2b} + b^{2a} <= 1 when a+b=1?

2010-03-05T18:15:16Z

This is too long to be a comment.

This inequality appears as conjecture 4.8 in this article here. As you probably know, V.Cirtoaje has written many books on olympiad-style inequalities, so you see my reason for not believing that a simple solution exists. Optimization problems can sometimes (or most of the time actually) require "non-elegant" analysis (whatever that means to you) so this search is a bit pointless in my opinion. If an elegant solution is found to some nontrivial optimization/estimation problem then it is very likely to appear in an olympiad/competition, and AOPS is the right place to carry such discussions.

Answer by Yaakov Baruch for Is there a good reason why a^{2b} + b^{2a} <= 1 when a+b=1?

2010-03-15T22:57:28Z

UPDATE: this "proof" is WRONG!
We want to prove that a^(2-2a)+(1-a)^(2a)<=1 for 0<=a<=1, or for 0<=a<=1/2 because of symmetry under a -> 1-a. Set f(a)=a^(2-2a). Then we want to prove that f(a)<=1-f(1-a), but since trivially f(a)<=a in [0,1/2], we have f(a)<=a<=1-a<=1-f(1-a). QED

Answer by fedja for Is there a good reason why a^{2b} + b^{2a} <= 1 when a+b=1?

2010-04-12T17:17:08Z

Fixed now. I spent some time looking for some clever trick but the most unimaginative way turned out to be the best. So, as I said before, the straightforward Taylor series expansion does it in no time.

Assume that $a>b$. Put $t=a-b=1-2b$.

Step 1: $$ \begin{aligned} a^{2b}&=(1-b)^{1-t}=1-b(1-t)-t(1-t)\left[\frac{1}2b^2+\frac{1+t}{3!}b^3+\frac{(1+t)(2+t)}{4!}b^4+\dots\right] \\ &\le 1-b(1-t)-t(1-t)\left[\frac{b^2}{1\cdot 2}+\frac{b^3}{2\cdot 3}+\frac{b^4}{3\cdot 4}+\dots\right] \\& =1-b(1-t)-t(1-t)\left[b\log\frac 1{a}+b-\log\frac {1}a\right] \\ &=1-b(1-t^2)+(1-b)t(1-t)\log\frac{1}a=1-b\left(1-t^2-t(1+t)\log\frac 1a\right) \end{aligned} $$ (in the last line we rewrote $(1-b)(1-t)=(1-b)2b=b(2-2b)=b(1+t)$)

Step 2. We need the inequality $e^{ku}\ge (1+u)(1+u+\dots+u^{k-1})+\frac k{k+1}u^{k+1}$ for $u\ge 0$. For $k=1$ it is just $e^u\ge 1+u+\frac{u^2}{2}$. For $k\ge 2$, the Taylor coefficients on the left are $\frac{k^j}{j!}$ and on the right $1,2,2,\dots,2,1$ (up to the order $k$) and then $\frac{k}{k+1}$. Now it remains to note that $\frac{k^0}{0!}=1$, $\frac{k^j}{j!}\ge \frac {k^j}{j^{j-1}}\ge k\ge 2$ for $1\le j\le k$, and $\frac{k^{k+1}}{(k+1)!}\ge \frac{k}{k+1}$.

Step 3: Let $u=\log\frac 1a$. We've seen in Step 1 that $a^{2b}\le 1-b(1-t\mu)$ where $\mu=u+(1+u)t$. In what follows, it'll be important that $\mu\le\frac 1a-1+\frac 1a t=1$ (we just used $\log\frac 1a\le \frac 1a-1$ here.

We have $b^{2a}=b(a-t)^t$. Thus, to finish, it'll suffice to show that $(a-t)^t\le 1-t\mu$. Taking negative logarithm of both sides and recalling that $\frac 1a=e^u$, we get the inequality $$ tu+t\log(1-te^u)^{-1}\ge \log(1-t\mu)^{-1} $$ to prove. Now, note that, according to Step 2, $$ \begin{aligned} &\frac{e^{uk}}k\ge \frac{(1+u)(1+u+\dots+u^{k-1})}k+\frac{u^{k+1}}{k+1} \ge\frac{(1+u)(\mu^{k-1}+\mu^{k-2}u+\dots+u^{k-1})}k+\frac{u^{k+1}}{k+1} \\ &=\frac{\mu^k-u^k}{kt}+\frac{u^{k+1}}{k+1} \end{aligned} $$ Multiplying by $t^{k+1}$ and adding up, we get $$ t\log(1-te^u)^{-1}\ge -ut+\log(1-t\mu)^{-1} $$ which is exactly what we need.

The end.

P.S. If somebody is still interested, the bottom line is almost trivial once the top line is known. Assume again that $a>b$, $a+b=1$. Put $t=a-b$.

$$ \begin{aligned} &\left(\frac{a^b}{2^b}+\frac{b^a}{2^a}\right)^2=(a^{2b}+b^{2a})(2^{-2b}+2^{-2a})-\left(\frac{a^b}{2^a}-\frac{b^a}{2^b}\right)^2 \\ &\le 1+\frac 14\{ [\sqrt 2(2^{t/2}-2^{-t/2})]^2-[(1+t)^b-(1-t)^a]^2\} \end{aligned} $$ Now it remains to note that $2^{t/2}-2^{-t/2}$ is convex on $[0,1]$, so, interpolating between the endpoints, we get $\sqrt 2(2^{t/2}-2^{-t/2})\le t$. Also, the function $x\mapsto (1+x)^b-(1-x)^a$ is convex on $[0,1]$ (the second derivative is $ab[(1-x)^{b-2}-(1+x)^{a-2}]$, which is clearly non-negative). But the derivative at $0$ is $a+b=1$, so $(1+x)^b-(1-x)^a\ge x$ on $[0,1]$. Plugging in $x=t$ finishes the story.

Answer by Will Jagy for Is there a good reason why a^{2b} + b^{2a} <= 1 when a+b=1?

2010-04-12T22:28:31Z

I think it gives a better sense of the geometry of the problem to ask whether, with non-negative $x,y$ such that $$ \frac{1}{2} \leq x + y \leq 1, $$ we can prove that $$ x^{2 y} + y^{2 x} \leq 1 ?$$ I'm not entirely certain where the second level curve component, through $\left( \frac{1}{4} , \frac{1}{4}\right),$ meets the axes. My programmable calculator seems to think that, if this arc does have $\left( \frac{1}{2} , 0 \right)$ as a limit point, the arc is tangent to the $x$-axis.

I see, this was pointed out in a comment on March 17 by Yaakov Baruch, one needs to click on the "show 6 more comments." I think I will leave this here anyway.

Answer by unknown (google) for Is there a good reason why a^{2b} + b^{2a} <= 1 when a+b=1?

2012-05-27T14:34:55Z

This type of problem can be solved by the following approach:

Maximize the function a^(2b)+b^(2a) s.t. a+b=1.
We find that the function is maximized at a=b=1/2 and takes value 1.

Answer by jbc for Is there a good reason why a^{2b} + b^{2a} <= 1 when a+b=1?

2013-05-12T09:45:39Z

Since this question has been bumped up I would like to state what I think is its natural framework: We have the inequality $f(a,b) \leq 1$ for $a+b=k$ when $k$ lies between $\frac 12$ and $1$. Otherwise, the inequality is $f(a,b) \leq \frac {k^k}{2^{k-1}}$ (with $f(a,b) = a^{2b}+b^{2a}$). This version is not just more comprehensive but it illustrates the dichotomy in where the maximum occurs (at the symmetric point $(\frac k2,\frac k2)$ or at the boundary $(k,0)$). The two cases considered above ($k=\frac 12$ and $k=1$) are precisely the transitional ones. One can also get estimates from below (usually by the constant $1$ but in a small neighbourhood around the critical interval $[\frac 12,1]$ the sharp version involves values which are given implicitly as the solution of transcendental equations).

(P.S. I had already given some of this information in a comment but, since it elicited no reaction, I have taken the liberty to repeat it here despite the fact that it isn't really an answer but, hopefully, does shed some light on the problem and its solution).