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.
 A: $$\Delta \le \frac {\sqrt{abc(a+b+c)}}{4}$$
$$\implies \Delta \le \frac {\sqrt{(4srR)(a+b+c)}}{4}$$
$$\implies\Delta \le \frac {\sqrt{(4\Delta R)(2s)}}{4}$$
$$\implies \sqrt{\Delta}\le\sqrt{\frac{Rs}{2}}$$
$$\implies \sqrt{sr}\le\sqrt{\frac{Rs}{2}}$$
$$\implies 2r\le R$$
$$\text{Which is obvious}$$
$s-\text{semi perimeter}, r-\text{inradius}, R-\text{circumradius}$
The equality in $2r\le R$ occurs when $a=b=c$ as circumcentre and incentre of equilateral triangle coincide.
I think it is necessary to add why $2r\le R $ obvious. It is due to this theorem:

If distance between Circumcentre and incentre of a triangle is $d$ then $d^2=R^2-2Rr$.

Since $d^2\ge0$ always so $R^2-2Rr\ge0\implies R\ge 2r$
A: By Heron's formula
$$\Delta=\sqrt{p(p-a)(p-b)(p-c)}$$
where $p=\frac{1}{2}(a+b+c)$.
Hence, the inequality is equivalent to
$$p(p-a)(p-b)(p-c) \le abcp$$
Note that, by the triangle inequality, $p\leq a+b$, $p\leq b+c$ and $p\leq a+c$.
Can you take it from here?
A: \begin{eqnarray*}
(a+b-c)(a-b)^2+(a-b+c)(a-c)^2+(-a+b+c)(c-b)^2 \geq 0
\end{eqnarray*}
The first bracket in each of the above is positive by the triangle inequality and the second bracket is a square, so the above quantity is clearly nonnegative. This can be rearranged to
\begin{eqnarray*}
(a+b-c)(a-b+c)(-a+b+c) \leq abc
\end{eqnarray*}
and the result now follows by using Heron's formula. It is also clear from the first inequality that equality will occur when $a=b=c$.
A: 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, $\Delta \leq \frac{1}{4}\sqrt{abc(a+b+c)}$.
