# Squaring inequality $a>b+c$

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

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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 $a$ $>$ $b$ $+$ $c$ $\implies$ $a^2$ $>$ $ab$ $+$ $ca$ $>$ $b^2$ $+$ $c^2$