Consider a,b,c are three side of a triangle. Now we need to find the relation between square the them sum of any two side of triangle with respect to third side.
My approach
As we know that the sum of any two sides of triangle will always be greater then the third side i.e
$(a+b)>c $
$(b+c)>a$
$(a+c)>b$
Take square on both side then
$a^2+b^2+2ab>c^2$
$ c^2+b^2+2bc>a^2$
$a^2+c^2+2ac>b^2$
Add last three equations
$2a^2+2b^2+2c^2-a^2-b^2-c^2>-2(ab+bc+ac)$
$a^2+b^2+c^2>-2(ab+bc+ac)$
Edited :
Corrected equation
But the correct answer is this $a^2 + b^2 + c^2 < 2(ab + bc + ca)$
What am I doing wrong?Please guide.
Best Answer
Repeat your reasoning starting with this form of triangular inequalities: $$ |a-b|<c,\quad |b-c|<a,\quad |c-a|<b, $$ that is: $$ (a-b)^2<c^2,\quad (b-c)^2<a^2,\quad (c-a)^2<b^2. $$