Let a , b and c be positive real numbers such that a + b + c = 3 . Find the minimum value of a a + b b + c c
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For any positive real number x , there are two possible cases -
{ If x ≥ 1 , then x m ≥ x n for any positive reals m > n . So x x ≥ x , when m = x , n = 1 . If x < 1 , then x m < x n for any positive reals m > n . So x < x x , when m = 1 , n = x .
In both cases we get x x ≥ x for all positive real numbers x .
∴ a a + b b + c c ≥ a + b + c = 3 ⟹ a a + b b + c c ≥ 3
Equality occurs when a = b = c = 1
Can I put this in the question itself as a follow up problem?
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Let f(x) = x^x. We can see that the graph of x^x is always positive having a minima at x =1/e. Basically it is a convex graph. Using Jensen's inequality for a convex graph , we have (f(a) + f(b) + f(c))/3 >= f( (a+b+c)/3). Equality holds when a=b=c. Thus a^a + b^b + c^c} >= 3f(1) as a+b+c=3. f(1)=1^1=1. Hence required answer is 3.