Question

If $a$, $b$, $c$ are positive real numbers and sides of the triangle, then prove that ${(a+b+c)}^3\ge 27(a+b-c)(c+a-b)(b+c-a)$

Answer 1

Let, $a$, $b$ and $c$ are positive real numbers and sides of the triangle. Then from the rule of forming triangle we have,

$(a+b>c),(c+a>b),(b+c>a)$

$\Rightarrow (a+b-c>0),(c+a-b>0),(b+c-a>0)$.

Now we take the positive real numbers $(a+b-c),(c+a-b),(b+c-a)$ and apply

$A.M.\ge G.M.$

$\Rightarrow \frac{(a+b-c)+(c+a-b)+(b+c-a)}{3}$ $\ge \sqrt[3]{3(a+b-c)(c+a-b)(b+c-a)}$

$\Rightarrow \frac{(a+b+c)}{3}$ $\ge \sqrt[3]{3(a+b-c)(c+a-b)(b+c-a)}$

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

$\Rightarrow (a+b+c{)}^3$$\ge 27(a+b-c)(c+a-b)(b+c-a)$.