Don't factorize!
Let a , b be postive real numbers satisfying a + b = 1 . If x 1 , x 2 , x 3 , x 4 , x 5 are positive real numbers so that x 1 x 2 x 3 x 4 x 5 = 1 , find the minimum value of ( a x 1 + b ) ( a x 2 + b ) ( a x 3 + b ) ( a x 4 + b ) ( a x 5 + b ) .
Solve this by factoring is easy, but solving it by any other method is hard (to me). Do you have any ideas?
The answer is 1.00.
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2 solutions
[This is not a complete solution.]
Show that ∏ ( a i + b i ) ≥ ( n ∏ a i + n ∏ b i ) n .
Hence, the result follows.
What about the 'x'es?
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What about them?
I never said that a i = a . Use the appropriate substitution.
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Oh. Thanks. I'll try to look for the substitution.
By Holder's inequality , ( a x 1 + b ) 1 / 5 ( a x 2 + b ) 1 / 5 ( a x 3 + b ) 1 / 5 ( a x 4 + b ) 1 / 5 ( a x 5 + b ) 1 / 5 ≥ a ( x 1 x 2 x 3 x 4 x 5 ) 1 / 5 + b = a + b = 1 . Therefore, ( a x 1 + b ) ( a x 2 + b ) ( a x 3 + b ) ( a x 4 + b ) ( a x 5 + b ) ≥ 1 . Equality occurs when all the x i are equal to 1, so the minimum value is 1.