# Is there a good reason why $a^{2b} + b^{2a} \le 1$ when $a+b=1$?

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 $a^{2b} + b^{2a}\le 1$.

• Scott, I don't get it. This is pretty much exactly what I'd expect from the title. If I were to title this post, I would write "elementary inequality", but "simple" seems close. And the phrasing is polite enough. Mar 5, 2010 at 17:10
• That said, this question might do better on artofproblemsolving.com . I could certainly bulldoze through it if I had to but, if there is a nice answer, then it is probably more likely to be found by a Putnam fellow or IMO medalist than a professional mathematician. Mar 5, 2010 at 17:15
• @David: what I don't like about the problem is that the questioner clearly knows the answer, so I have no incentive to think about it myself in this forum, which I percieve to be for people who are "stuck" or "need help". There are a gazillion resources for putnam-like problems on the web, but this place seems to generally discourage it, which is something I like about it. Mar 5, 2010 at 18:18
• Regarding the title, I guess this is just a pet rant by now, but I wish the community norms were: 1) Always ask a question in your title and 2) try to make that question as close as possible an approximation to the full question you're asking. Unfortunately everyone uses the title more as an email subject line. Mar 5, 2010 at 18:36
• Rather than argue in comments I used my editing power to change the title to an actual question. This is the most common sort of edit I make to people's comments and I think it's a good thing to just do. Apr 13, 2010 at 1:58

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.

• Bravo ! Apr 13, 2010 at 4:26
• @fedja: I think your proof meets the criteria. If you are interested, you may submit your proof featuring solving a conjecture in <On Some Inequalities With Power-Exponential Functions> by Vasile Cirtoaje. Apr 13, 2010 at 13:26
• Thanks. Well, anybody who is interested can find it here, so I do not think it makes much sense to submit it anywhere else. I'll, probably, just send a PM to Vasile. Apr 13, 2010 at 14:26
• @fedja: That is good. For me, I learned a good proof, that is all I want for posting this problem. Apr 13, 2010 at 15:22
• With respect, I would suggest that the following form of the inequality is more interesting for several reasons: $f(a,b)\leq \frac{k^k}{2^{k-1}}$ if $a+b=k$ ($f$ denotes, of course, the expression to be maximised It includes the cases $a+b=\frac 12$ and $a+b=1$, is more precise and displays more interesting behaviour (the dependence of the maximum on $k$ and the nature of the point where it is assumed.
– jbc
Nov 27, 2012 at 17:41

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.

• It now becomes a problem whether I should post this sort of problem here (my interest in this problem is simple: curiosity). Yes, generally problems from olympiad-style is not welcome here. I expect that new and fresh ideas may appear here. Mar 5, 2010 at 18:26
• An argument that backs this up to some extent is the fact that the maximum occurs at (0,1), (1/2,1/2) and (1,0). What's more, the place where the minimum occurs is, if my calculation is correct, the place where aloga = 1-a, which doesn't fill one with confidence that a slick solution exists. Even so, I don't completely rule it out. Mar 5, 2010 at 18:30
• Wait, this was actually published as a conjecture? I'm certain this inequality is provable if you let me use a computer. For example, let $f[t]=t^{2(1−t)}$. Then, for $k/10000 \leq t \leq (k+1)/10000$ , we have $f[t]+f[1−t] \leq f[(k+1)/10000]+f[1−k/10000]$ . This proves the inequality for all $t$ not in $[0,3/10000]$ , $[4747/10000, 5253/10000]$ and $[9997/10000,1]$. Local arguments near 0, $1/2$ and 1 should finish the job. This is only a hard problem if you insist on a simple, non-machine-aided answer. Mar 5, 2010 at 18:39
• @DS: I'm sure that's what the author meant by stating it as a "conjecture", that there is no solution similar to the simple proofs of the other inequalities mentioned in the article. I just gave it as a counterargument to miwalin insisting that we find an elegant solution here. Mar 5, 2010 at 18:41
• @GZ: If I were the referee, I would insist on the author distinguishing results that don't have proofs from results that don't have elementary proofs. But in any case, nice work finding that reference! That definitely makes it seem less likely that there is an elegant approach. Mar 5, 2010 at 18:48

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).

By symmetry, it is enough to show that $$f(x):=x^{2-2 x}+(1-x)^{2 x}\le1$$ for $$x\in(0,\tfrac12);$$ the latter condition will be assumed by default. It is easy to see that $$r_0(x):=1 - x - x \ln x>0.$$ So, $f'(x)$ equals $$f_1(x):=\frac{f'(x)}{x^{1 - 2 x} r_0(x)}$$ in sign. Moreover, $$f_2(x):=f'_1(x)\frac{r_0(x)^2}{2 \big((1 - x) x\big)^{2 x}}$$ is a rational function of $x,\ln x,\ln(1-x)$, and $f_2(x)$ equals $f'_1(x)$ in sign.

This almost completes the solution, since the sign pattern of such a rational function can be determined completely algorithmically, and then one can go back and determine the sign patterns of $f_1$ and $f-1$, in this order. So, the only remaining problem is to find a better algorithm.

Differentiating $f_2(x)$, we kill the terms containing $\ln^2 x\ln(1-x)$ and $\ln x\ln^2(1-x)$, so that $$f_3(x):=f'_2(x)/c_3(x)$$ is a polynomial of degree 1 in $\ln x,\ln^2 x,\ln(1-x),\ln^2(1-x),\ln x\ln(1-x)$ (with coefficients in $\mathbb Z(x)$), where $$c_3(x):=\frac{4 \left(2 x^3-3 x^2-x+1\right)}{(1-x)^2 x^2}>0.$$ Next, $$f_4(x):=f'_3(x)\,\frac{(1 - x)^3 x^3 c_3(x)^2}{4 - x + x^2}$$
is a polynomial of degree 1 in $\ln x,\ln^2 x,\ln(1-x),\ln^2(1-x)$ and $$f_5(x):=f'_4(x)\,\frac{x^2 (4 - x + x^2)^2}{2c_3(x)(1-x)c_5(x)}$$
is a polynomial of degree 1 in $\ln x,\ln(1-x)$, where $c_5(x):=12 + 4 x + 47 x^3 - 99 x^4 + 10 x^5 - 6 x^6>0$.

Further, $$f_6(x):=f'_5(x)\, \frac{c_5(x)^2 (1 - x)^4}{x (4 - x + x^2)}$$
is a polynomial of degree 1 in $\ln x$, and hence so is $f'_6(x)$. Therefore, the value of $f'_6(x)$ is between those of $f_{73}(x)$ and $f_{74}(x)$, where $f_{7k}(x)$ is obtained from $f'_6(x)$ by replacing there $\ln x$ with the Taylor polynomial of degree $k$ for $\ln x$ at the point $1/2$. Since $f_{73}(x)$ and $f_{74}(x)$ are rational expressions (in $\mathbb R(x)$), it is easy to see that $f_{73}<0$, $f_{74}<0$, and hence $f'_6<0$, so that $f_6$ decreases (on $(0,1/2)$). Moreover, $f_6(0+)>0>f_6(1/2)$. So, $f_6$ is $+-$; that is, it changes in sign only once on $(0,1/2)$, from $+$ to $-$.

So, $f_5$ is up-down; that is, it increases on $(0,c)$ and decreases on $(c,1/2)$, for some $c\in(0,1/2)$. Also, $f_5(0+)=0=f_5(1/2)$. So, $f_5>0$, $f_4$ increases, to $f_4(1/2)=0$. So, $f_4<0$, $f_3$ decreases, from $f_3(0+)=0$. So, $f_3<0$, $f_2$ decreases, from $f_2(0+)=\infty$ to $f_2(1/2)=-1<0$. So, $f_2$ is $+-$, $f_1$ is up-down, with $f_1(0+)=-2<0$ and $f_1(1/2)=0$. So, $f_1$ is $-+$, $f$ is down-up, with $f(0+)=f(1/2)=1$. So, $f<1$ on $(0,1/2)$. QED

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.

• Since $e^{at}t^t$ is convex for all $a\in\mathbb R$ (a boring but straightforvard computation shows that the second derivative is $[(a+\log t+1)^2+\frac 1t]e^{at}t^t$), including the point $(x,y)$ into the family $(xt,yt)$, we see that it is enough to check the boundary curves. I've done the upper one already. Anybody wants to try the lower one? Apr 13, 2010 at 4:35

This answer implements a very head on, non creative approach to the problem, as mentioned in David Speyer's comment to Gjergji Zaimi's answer. I thought it's worth having it here, but I regret unnecessarily bumping up the thread because of it.

Assume $a\le\frac{1}{2}$ and set $a=\frac{1}{x}$ with $x\ge 2$; then the inequality becomes

$1-\frac{1}{x} \stackrel{?}{\le}\left(1-x^{2/x-2}\right)^{x/2}$

If $x\ge 2$ then $\text{RHS}\ge 1-x^{2/x-2}\cdot x/2=1-\frac{x^{2/x}}{2x}$ and in turn $1-\frac{x^{2/x}}{2x}\ge1-\frac{1}{x}\Longleftrightarrow x^2\ge2^x$, which is esily seen to hold if $x\ge 4$, or $a\le1/4$.

For $a$ near $1/2$, substitute $2a=1-\epsilon$ and $2b=1+\epsilon$, so that $b^{2a}+a^{2b}\le1$ becomes $2^{\epsilon}(1+\epsilon)^{1-\epsilon}+2^{-\epsilon}(1-\epsilon)^{1+\epsilon}\le 2 \tag{1}$

$2^{\epsilon}(1+\epsilon)^{1-\epsilon}+2^{-\epsilon}(1-\epsilon)^{1+\epsilon}=\big(1+l\epsilon+\frac{l^2}{2!}\epsilon^2\cdots \big)\big(1+\epsilon{1-\epsilon\choose 1}+\epsilon^2{1-\epsilon \choose 2}\cdots\big) +\big(1-l\epsilon+\frac{l^2}{2!}\epsilon^2\cdots \big) \big(1-\epsilon{1+\epsilon\choose 1}+\epsilon^2{1+\epsilon \choose 2}\cdots\big)$

where $l=\log(2)=0.693147...$

Clearly $\big(1\pm l\epsilon+\frac{l^2}{2!}\epsilon^2\cdots \big)\le\big(1\pm l\epsilon+\frac{l^2}{2!}\epsilon^2+\frac{l^3}{6}\cdot\frac{\epsilon^3}{1-\epsilon} \big)<\big(1\pm l\epsilon+\frac{l^2}{2!}\epsilon^2+\frac{1}{18}\cdot\frac{\epsilon^3}{1-\epsilon} \big)$

and similarly

$\big(1\pm\epsilon{1\mp\epsilon\choose 1}+\epsilon^2{1\mp\epsilon \choose 2}\cdots\big) \le \big(1\pm\epsilon{1\mp\epsilon\choose 1} + \epsilon\cdot \frac{\epsilon^2}{1-\epsilon}\big)=\big(1\pm\epsilon +\epsilon^2 + \frac{\epsilon^3}{1-\epsilon}\big)$

because ${1\mp\epsilon \choose n}=\frac{1\mp\epsilon}{2}\cdot\frac{\mp \epsilon}{1}\cdot\frac{\mp \epsilon-1}{3}\cdot\frac{\mp \epsilon-2}{4}\cdots\le 1\cdot\epsilon\cdot 1\cdot 1\cdots=\epsilon$, if $n\ge2$.

So

$2^{\epsilon}(1+\epsilon)^{1-\epsilon}+2^{-\epsilon}(1-\epsilon)^{1+\epsilon}<\big(1+ l\epsilon+\frac{l^2}{2!}\epsilon^2+\frac{1}{18}\cdot\frac{\epsilon^3}{1-\epsilon} \big)\big(1+\epsilon +\epsilon^2 + \frac{\epsilon^3}{1-\epsilon}\big)+\big(1- l\epsilon+\frac{l^2}{2!}\epsilon^2+\frac{1}{18}\cdot\frac{\epsilon^3}{1-\epsilon} \big)\big(1-\epsilon +\epsilon^2 + \frac{\epsilon^3}{1-\epsilon}\big)=-(\frac{1}{9}\cdot\frac{\epsilon}{1-\epsilon}+l^2)(1-\frac{\epsilon}{1-\epsilon})\cdot \epsilon^4+(\frac{19}{9}\cdot\frac{\epsilon}{1-\epsilon}+l^2+2l-2)\cdot \epsilon^2+2$

now $\text{RHS}< (\frac{19}{9}\cdot\frac{\epsilon}{1-\epsilon}+l^2+2l-2)\cdot \epsilon^2+2<2\space$ if $\space \epsilon\le 0.05$, which means that the original inequality is true for $a\le0.25$ and for $0.475\le a \le 0.5$.

Last, since $(1-a)^{2a}$ is decreasing and $a^{2-2a}$ increasing, one completes the proof by verifying that $(1-a_i)^{2a_i}+a_{i+1}^{2-2a_{i+1}} \le1$ for $a_i=0.25+i/200\space$ for $\space 0\le i \le 44$.

We want to show 1:

Let $$0 such that then we have : $$f(x)=x^{2(1-x)}+(1-x)^{2x}\leq q(x)=(1-x)^{2x}+2^{2x+1}(1-x)x^2\leq 1$$

The Lhs is equivalent to :

$$x^{2(1-x)}\leq h(x)=2^{2x+1}(1-x)x^2$$

Or : $$\ln\Big(x^{2(1-x)}\Big)\leq \ln\Big(2^{2x+1}(1-x)x^2\Big)$$

Making the difference of these logarithm and introducing the function :

$$g(x)=\ln\Big(x^{2(1-x)}\Big)-\ln\Big(2^{2x+1}(1-x)x^2\Big)$$

The derivative is not hard to manipulate and we see that it's positive and $$x=0.5$$ is an extrema .The conclusion is :

$$g(x)\leq g(0.5)=0$$

And we are done with the LHS.

For the Rhs I use one of the lemma (7.1) due to Vasile Cirtoaje we have :

$$(1-x)^{2x}\leq p(x)=1-4(1-x)x^{2}-2(1-x)x(1-2 x)\ln(1-x)$$

So we have :

$$q(x)\leq p(x)+h(x)$$

We want to show that :

$$p(x)+h(x)\leq 1$$

Wich is equivalent to :

$$-2(x-1)x((4^x-2)x+(2x-1)\ln(1-x))\leq 0$$

It's not hard so I omitt here the proof of this fact .

We are done .

1 Vasile Cirtoaje, "Proofs of three open inequalities with power-exponential functions", The Journal of Nonlinear Sciences and its Applications (2011), Volume: 4, Issue: 2, page 130-137. https://eudml.org/doc/223938

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.
• Did you try plugging in $a=0,b=1$? May 28, 2012 at 20:07