Question

Show that in any triangle with sides a, b and c, we have $(a+b+c{)}^2<4(ab+bc+ca).$

Answer 1

Without loss of generality, let us assume that a > b > c > O. Since the sum of any two sides of a triangle is always greater than the third side, therefore the difference between any two sides can never exceed the third and hence

$0\le b-c<a,0\le c-a<b,0\le a-b<c$

Squaring both sides of the above inequalities and adding the corresponding sides,

We get
$(b-c{)}^2+(a-c{)}^2+(a-b{)}^2<a^2+b^2+c^2$
$\iff a^2+b^2+c^2<2(bc+ca+ab)$
$\iff (a+b+c{)}^2<4(bc+ca+ab)$