RMO 2012 solution to Question No. 3

3. Let a and b are positive real numbers such that a+b = 1. Prove that \(a^a  b^b + a^b b^a \le 1\)

Solution:

We use the weighted A.M.-G.M. inequality which states that:

\(\frac {w_1 a_1 + w_2 a_2 }{w_1 + w_2} \ge ({a_1}^{w_1} {a_2}^{w_2})^{\frac{1}{w_1 + w_2}} \)

First we put \(w_1 = a , a_1 = a , w_2 = b, a_2 = b \)
Hence we get \(\frac {a a + b b }{a + b} \ge (a^a b^b)^{\frac{1}{a + b}} \)
As a+b =1
we have \(a^2 + b^2 \ge (a^a b^b) \) ....(1)

Similarly we put \(w_1 = a , a_1 = b , w_2 = b, a_2 = a \)

Hence we get \(\frac {a b + b a }{a + b} \ge (b^a a^b)^{\frac{1}{a + b}} \)
As a+b =1
we have \(2ab \ge (b^a a^b) \) ....(2)

Adding (1) and (2) we have 

\(a^2 + b^2 + 2ab \ge a^a b^b + b^a a^b \)
=> \((a+b)^2 \ge a^a b^b + b^a a^b \)
As a+b =1 we have the desired inequality
\(1 \ge a^a b^b + b^a a^b \).