Question

Prove that the difference of any two sides is less than Third side

Answer 1

Let a , b and c are the sides of traingle ABC,
$a \ \textgreater \ 0 \\ b \ \textgreater \ 0 \\ c \ \textgreater \ 0$
Hence, AM > GM
$\frac {b+c}{2} \geq \sqrt {bc}$
$CosA = \frac {b^2 + c^2 - a^2 }{2bc} \\ 0 \ \textless \ A \ \textless \ 180$
After solving this,we get
$b + c \ \textgreater \ a$
Similarly,$c + a \ \textgreater \ b \\ a + b \ \textgreater \ c$
Hence, sum of two sides, of any traingle is always greater then,
$hence,\\ b \ \textless \ a - c \\ c \ \textless \ b - a \\ a \ \textless \ c - b$

Answer 2

Hi there ♦♦♦♦♦♦


Theorem which is, the sum of any two sides of a triangle must be greater than the third. Stated algebraically: 
A < B + C 
B < A + C 
C < A + B 

1. Subtract B from both sides, then A - B < C 
2. Subtract C from both sides, then B - C < A 
3. Subtract A from both sides, then C - A < B


Hope it is helpful ans.♦♦♦♦

Thanks ♥♥♥