Smooth again

Algebra Level 3

Given non-negative real numbers x 1 , x 2 , … , x n x_{1},x_{2},\ldots,x_{n} satisfying x 1 + x 2 + ⋯ + x n ≤ 1 2 , x_{1}+x_{2}+\cdots+x_{n}\le\frac{1}{2}, what is the minimum value of ( 1 − x 1 ) ( 1 − x 2 ) ⋯ ( 1 − x n ) ? (1-x_{1})(1-x_{2})\cdots(1-x_{n})?


The answer is 0.5.

Haosen Chen by Haosen Chen , 20, China

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1 solution

Pedro Cardoso
May 10, 2018

Consider the product ( 1 − a ) ( 1 − b ) (1-a)(1-b) where a + b = k ≤ 1 2 a+b=k\le \frac{1}{2} and a , b > 0 a,b > 0 .

Since 1 − a − b + a b > 1 − a − b 1-a-b+ab>1-a-b We can see that ( 1 − a ) ( 1 − b ) > ( 1 − 0 ) ( 1 − a − b ) (1-a)(1-b)>(1-0)(1-a-b)

This means if we have a product ( 1 − x a ) ( 1 − x b ) (1-x_a)(1-x_b) , we can decrease the product without altering the sum by replacing x a x_a with 0 0 and x b x_b with x a + x b x_a+x_b

By doing this repeatedly over all x n x_n , we have

( 1 − ( x 1 + x 1 + . . . + x n ) ) = ( 1 − 1 2 ) = 1 2 (1-(x_1+x_1+...+x_n))=(1-\frac{1}{2})=\frac{1}{2}

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