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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
$(1 \ge a^a b^b + b^a a^b )$.