<s>[INST] 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
 [/INST] 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 $a$ $>$ $b$ $+$ $c$ $\implies$ $a^2$ $>$ $ab$ $+$ $ca$  $>$ $b^2$ $+$ $c^2$
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