Calculate the inequality

Question

$a,b,c$ are the sides of a triangle.Then show that

$$(a+b+c)^3 > 27(a+b-c)(b+c-a)(c+a-b)$$ Also give the case where equality holds i.e. $$(a+b+c)^3=27(a+b-c)(b+c-a)(c+a-b)$$

I tried triangle inequality, AM-GM inequality in many way but cannot help my cause.

Answer 1

First apply AM-GM inequality: $(a+b+c)^3 \geq 27abc$, then you need to prove: $abc \geq (a+b-c)(b+c-a)(c+a-b)$, and this follows from:

$a^2 \geq a^2 - (b-c)^2$

$b^2 \geq b^2 - (c-a)^2$

$c^2 \geq c^2 - (a-b)^2$.

by multiply all three inequalities and taking square root.

Answer 2

Using $s=\frac{a+b+c}2$, this transforms to:

$$s^3\geqslant27(s-a)(s-b)(s-c)$$

Which is true by AM GM,

$$[(s-a)(s-b)(s-c)]^{1/3}\leqslant\frac{3s-a-b-c}{3}=\frac s 3$$

Answer 3

By AM-GM $$(a+b+c)^3=\left(\sum_{cyc}a\right)^3=\left(\sum_{cyc}(a+b-c)\right)^3\geq$$ $$\geq\left(3\sqrt[3]{\prod_{cyc}(a+b-c)}\right)^3=27(a+b-c)(a+c-b)(b+c-a).$$ Done!