Question

If $a,b,c$ are sides of a scalene triangle then prove that$(a+b+c{)}^3>27$ $(a+b-c)(b+c-a)(c+a-b)$

Answer 1

$2s=a+b+c$ $\Rightarrow$ $a+b-c=2s-2c$

Similarly $a-b+c=2s-2b,b+c-a=2s-2a$
(Using AM > GM, a $\ne$ b $\ne$c )

$\frac{2s-2c+2s-2a+2s-2b}{3}>\left[\right(a+b-c\left)\right(b+c-a\left)\right(c+a-b){]}^{1/3}$

$\Rightarrow \frac{2s}{3}>\left[\right(a+b-c\left)\right(b+c-a\left)\right(c+a-b){]}^{1/3}$

$\Rightarrow (a+b+c)>3\left[\right(a+b-c\left)\right(b+c-a\left)\right(c+a-b){]}^{1/3}$

$\Rightarrow (a+b+c{)}^3>27(a+b-c\left)\right(b+c-a\left)\right(c+a-b)$