$LHS\leq a^2+b^2+2ab=1$ , just AM-GM :O

yes, he is right. by using weightd AM.GM , we get the required result.

Using the expansion, (a+b)^(a+b)

we can also use bimomial theorem to solve it.

yeah @azaza

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?

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.

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.

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.

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

By Weighted AM-GM we get \[a^ab^b\leq{a^2+b^2},a^bb^a\leq2ab\] so, \[{a^ab^b+a^bb^a}\leq{a^2+b^2+2ab}=1\]

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?

A similar problem is there in Samin Riasat's inequalities e-book.

Believe me or not, RMO 2012 Region 1 paper was very,very easy.

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.

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 $1$.$Q.E.D.$.....

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.

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$

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

$$a^ab^b +a^bb^a \leq \frac{a\cdot a + b\cdot b}{a+b} + \frac{b \cdot a + a \cdot b}{a+b} \leq a^2 + b^2 + 2ab = (a+b)^2 = 1$$