2018/001) Show that if a,b,c are sides of a triangle then $(a+b+c)^3 >= 27(a+b-c)(b+c-a) ( c + a - b)$

We have
$a^2 >= a^2 - (b-c)^2$ or $a^2 >= (a+b-c)(a-b + c) \cdots(1)$
Similarly $b^2 >= (b+c-a)(b-c+a)\cdots(2)$
and  $c^2 >= (c+a-b)(c-a+b)\cdots(3)$

Multiply (1) , (2) , (3) to get
$(abc)^2 >= ((a+b-c) (b+c-a)(c+a-b))^2$
or
$abc > = (a+b-c)(b+c-a)(c+a-b)\cdots(4)$

Applying $AM >= GM$ to $a,b,c$ we get

$\frac{a+b+c}{3} >= \sqrt[3]{abc}$

or $(a+b+c)^3 >= 27abc\cdots(5)$

From (4) and (5) we get

$(a+b+c)^3 >= 27(a+b-c)(b+c-a) ( c + a - b)$