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December 3, 2012

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

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