Let $a$ and $b$ be positive real numbers such that $a+b=1$. Prove that $a^ab^b+a^bb^a\le 1$.
Problem
Source: RMO 2012
Tags: logarithms, function, inequalities, inequalities unsolved
02.12.2012 13:40
$LHS\leq a^2+b^2+2ab=1$ , just AM-GM :O
02.12.2012 13:48
yes, he is right. by using weightd AM.GM , we get the required result.
02.12.2012 18:54
Using the expansion, (a+b)^(a+b)
03.12.2012 07:03
we can also use bimomial theorem to solve it.
03.12.2012 08:05
yeah @azaza
03.12.2012 13:24
I used weighted AM-GM. But, the problem is that weighted AM-GM is only defined when the weights are all rational. But here a and b need not necessarily be rational. What do you think guys? Will they give marks for this?
03.12.2012 13:36
The Weights in Weighted AM-GM can be anything as long as they are positive reals. Its not necessarily rationals. Your method is absolutely right and nobody should discredit your solution.
03.12.2012 15:22
Sampro wrote: I used weighted AM-GM. But, the problem is that weighted AM-GM is only defined when the weights are all rational. But here a and b need not necessarily be rational. What do you think guys? Will they give marks for this? weighted AM GM is extended to real numbers using the continuity argument.
03.12.2012 17:44
Adi2497 wrote: Sampro wrote: I used weighted AM-GM. But, the problem is that weighted AM-GM is only defined when the weights are all rational. But here a and b need not necessarily be rational. What do you think guys? Will they give marks for this? weighted AM GM is extended to real numbers using the continuity argument. Could you explain it a bit more? I just want to be sure.
04.12.2012 16:53
here is a sol using concavity, since logarithm function is concave, log(a^ab^b)+log(a^bb^a)by2>log(a^a+b^b by2)>log(2^-1) which gives ab<4^-1 but this is already true from am gm theorem
04.12.2012 20:11
By Weighted AM-GM we get \[a^ab^b\leq{a^2+b^2},a^bb^a\leq2ab\] so, \[{a^ab^b+a^bb^a}\leq{a^2+b^2+2ab}=1\]
07.12.2012 08:49
By AM-GM (normal, not weighted ) , we get ab <= 1/4. Equality holds iff a = b = 1/2. Can we make the generalisation that, if ab is minimum , the expression given in the problem is minimum ? If we substitute a = b = 1/2 in the expression, we do get 1 but is the method correct?
13.01.2013 09:42
A similar problem is there in Samin Riasat's inequalities e-book.
23.02.2014 21:37
Believe me or not, RMO 2012 Region 1 paper was very,very easy.
26.02.2014 20:18
Since it was already revived, i am going to post one more sol(mine) for this prob. $(a+b)^a(b+a)^b\geq (a^ab^b+b^aa^b)$ as $a+b=1$ by holder ineq.
23.09.2015 20:56
Another Solution using $RI$..Assume wlog that $a\geq b$. Now proceed with the Rearrangement Inequality to get $a^a.a^b+b^a.b^b$$\geq$$a^a.b^b+a^b.b^a$. The $LHS$ equals $1$.$Q.E.D.$.....
16.09.2024 17:51
We use the weighted AM-GM inequality twice. $(a^ab^b)^{\frac{1}{a+b}}\leq \frac{a^2+b^2}{a+b}$ And $(a^bb^a)^\frac{1}{a+b}\leq \frac{2ab}{a+b}$ Upon adding these two inequalities, we get that\hfill\break $a^ab^b+a^bb^a\leq \frac{(a+b)^2}{(a+b}=1$, and we are done.
16.09.2024 18:06
Super unoriginal but it works. Observe the following two chains of inequalities: $$a^ab^b = (a^ab^b)^{\frac{1}{a+b}} \le \frac{a^2 + b^2}{a+b} = a^2 + b^2$$$$a^bb^a = (a^bb^a)^{\frac{1}{a+b}} \le \frac{2ab}{a+b} = 2ab.$$both of which are by Weighted AM-GM. Adding, $$a^ab^b+a^bb^a\le a^2 + b^2 + 2ab = (a+b)^2 = 1.$$$\square$
16.09.2024 20:35
AshAuktober wrote: Super unoriginal but it works. Observe the following two chains of inequalities: $$a^ab^b = (a^ab^b)^{\frac{1}{a+b}} \le \frac{a^2 + b^2}{a+b} = a^2 + b^2$$$$a^bb^a = (a^bb^a)^{\frac{1}{a+b}} \le \frac{2ab}{a+b} = 2ab.$$both of which are by Weighted AM-GM. Adding, $$a^ab^b+a^bb^a\le a^2 + b^2 + 2ab = (a+b)^2 = 1.$$$\square$ See also here: here
21.09.2024 08:44
$$a^ab^b +a^bb^a \leq \frac{a\cdot a + b\cdot b}{a+b} + \frac{b \cdot a + a \cdot b}{a+b} \leq a^2 + b^2 + 2ab = (a+b)^2 = 1$$