Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Given $a,b,c > 0$ and $a > b + c$ , is it true that $a^2 > b^2 + c^2$?

Im tried proving it. I followed the below steps and not sure whether im right or wrong?

Squaring on both sides
$\implies a^2 > (b+c)^2$
$\implies a^2 > b^2 + c^2 + 2*b*c$
since $b,c > 0 $, we have $2*b*c > 0$, then
$\implies a^2 > b^2 + c^2$ Q.E.D

share|cite|improve this question
You are completely correct. – Souvik Dey Nov 7 '12 at 9:10
Think of 2 squares one with side a and the other with side b+c, you'd immediately see that the area of the of the square with the bigger side is the bigger area. – NoChance Nov 7 '12 at 10:22
It's easy to check your proof by thinking of it geometrically: draw a square with side b+c: the squares with sides b and c can be fitted with space left over, and the larger square fits inside the square of side a. – Charles Stewart Nov 7 '12 at 10:36

Another way of proving is ; since $a$ , $b$ , $c$ $>$ $0$ then $a$ $>$ $b$ $+$ $c$ implies $a$ $>$ $b$ and $a$ $>$ $c$ ; so $ab$ $>$ $b^2$ and $ca$ $>$ $c^2$ , and $a$ $>$ $b$ $+$ $c$ $\implies$ $a^2$ $>$ $ab$ $+$ $ca$ $>$ $b^2$ $+$ $c^2$

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.