inequality related to square the sum of any two sides of a triangle with respect to square of other side

Question

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.

Answer 1

Repeat your reasoning starting with this form of triangular inequalities: $$ |a-b|<c,\quad |b-c|<a,\quad |c-a|<b, $$ that is: $$ (a-b)^2<c^2,\quad (b-c)^2<a^2,\quad (c-a)^2<b^2. $$

Answer 2

The inequality you attained is trivial as the lengths of sides of a triangle can't be negative or zero. In fact the correct answer comes from$$(a-b)^2+(a-c)^2+(b-c)^2< c^2+b^2+a^2$$which is a direct outcome of triangle inequality you mentioned above (how?).