The square is never negative!

Algebra Level 2

Let a , b , c , d a,b,c,d be real numbers such that the numbers 2 a − b 2 , 2 b − c 2 , 2 c − d 2 2a-b^2, 2b-c^2, 2c-d^2 and 2 d − a 2 2d-a^2 are all greater than or equal to 1. Find a + b + c + d a+b+c+d .


The answer is 4.

Santi Spadaro by Santi Spadaro , 39, Italy

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2 solutions

Santi Spadaro
Aug 9, 2018

The inequalities 2 a − b 2 ≥ 1 , 2 b − c 2 ≥ 1 , 2 c − d 2 ≥ 1 , 2 d − a 2 ≥ 1 2a-b^2 \geq 1, 2b-c^2 \geq 1, 2c-d^2 \geq 1, 2d-a^2 \geq 1 all hold at the same time, so we can add them up, obtaining the inequality 2 a − b 2 + 2 b − c 2 + 2 c − d 2 + 2 d − a 2 ≥ 1 + 1 + 1 + 1 2a-b^2+2b-c^2+2c-d^2+2d-a^2 \geq 1+1+1+1 .

Moving everything to the right hand side the above inequality becomes a 2 − 2 a + 1 + b 2 − 2 b + 1 + c 2 − 2 c + 1 + d 2 − 2 d + 1 ≤ 0 a^2-2a+1+b^2-2b+1+c^2-2c+1+d^2-2d+1 \leq 0 , that is ( a − 1 ) 2 + ( b − 1 ) 2 + ( c − 1 ) 2 + ( d − 1 ) 2 ≤ 0 (a-1)^2+(b-1)^2+(c-1)^2+(d-1)^2 \leq 0 .

But this inequality is true if and only if a = b = c = d = 1 a=b=c=d=1 . So a + b + c + d = 4 a+b+c+d=4 .

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Edwin Gray
Sep 10, 2018

Add the 4 inequalities, resulting in 2a + 2b + 2c + 2d >= 4 + a^2 + b^2 + c^2 + d^2. Transposing, 0 >= (a - 1)^2 + (b - 1)^2 + (c - 1)^2 + (d - 1)^2. This can only happen if a = b = c = d = 1, so a + b + c + d = 4. Ed Gray

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