Show that $\Delta \le \frac {\sqrt{abc(a+b+c)}}{4}$ If $\Delta$ is the area of a triangle with side lengths a, b, c, then show that: $\Delta \le \frac {\sqrt{abc(a+b+c)}}{4}$. Also show that equality occurs in the above inequality if and only if a = b = c.
I am not able to prove the inequality.

One has $\Delta = \frac{1}{4}\sqrt{(a+b+c)(b+c-a)(c+a-b)(a+b-c)}$ (See this link).
Moreover, one has $(a+b-c)(b+c-a) \leq (\frac{a+b-c + b+c-a}{2})^2 = b^2$
So $[(a+b+c)(b+c-a)(c+a-b)(a+b-c)]^2 \leq a^2b^2c^2$.
Thus, $\Delta \leq \frac{1}{4}\sqrt{abc(a+b+c)}$.
