Quadro quadratic

Algebra Level 3

a , b , c a,b,c are real numbers. If a 2 + b 2 + c 2 = 1 a^2+b^2+c^2=1 , then a b + b c + c a ab+bc+ca lies in the interval:

[-1,2] [1/2,1] [-1/2,1] [-1,1/2]

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Devkant Chouhan by Devkant Chouhan , 23, India

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1 solution

Jordi Bosch
Nov 6, 2014

( a + b + c ) 2 ≥ 0 (a+b+c)^2\ge0 a 2 + b 2 + c 2 + 2 ( a b + b c + c a ) ≥ 0 a^2+b^2+c^2+2(ab+bc+ca)\ge0 − 1 2 ≤ a b + b c + c a -\frac{1}{2} \leq ab +bc +ca On the other hand, we have: 1 2 ( ( a − b ) 2 + ( b − c ) 2 + ( c − a ) 2 ) ≥ 0 \frac{1}{2}\left((a-b)^2 + (b-c)^2 + (c-a)^2\right)\ge0 a 2 + b 2 + c 2 ≥ a b + b c + c a a^2+b^2+c^2\ge ab+bc+ca 1 ≥ a b + b c + c a 1\ge ab + bc + ca

31 Helpful 0 Interesting 0 Brilliant 0 Confused

Nicely explained

Devkant Chouhan - 6 years, 7 months ago

What made you think of the second part of the solution? I came up with the first half and was able to guess my way to the answer, but is there any motivation behind the second part?

Gino Pagano - 6 years, 7 months ago

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Oh that is indeed a very well-known equation. The motivation behind it was that I already knew the finally inequality, it appears quite often.

Jordi Bosch - 6 years, 7 months ago

[-1,2] contains all other intervals, so it is correct as well

Diego Marcon Farias - 5 years, 8 months ago

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