proving the inequality $\triangle\leq \frac{1}{4}\sqrt{(a+b+c)\cdot abc}$ If $\triangle$ be the area of $\triangle ABC$ with side lengths $a,b,c$. Then show that $\displaystyle \triangle\leq \frac{1}{4}\sqrt{(a+b+c)\cdot abc}$ 
and also show that equality hold if $a=b=c$.
$\bf{My\; Try}::$ Here we have to prove $4\triangle\leq \sqrt{(a+b+c)\cdot abc}$
Using the formula $$\triangle = \sqrt{s(s-a)(s-b)(s-c)},$$ where $$2s=(a+b+c)$$
So $$4\triangle = \sqrt{2s(2s-2a)(2s-2b)(2s-2c)}=\sqrt{(a+b+c)\cdot(b+c-a)\cdot(c+a-b)\cdot(a+b-c)}$$
Now using $\bf{A.M\geq G.M}$ for $(b+c-a)\;,(c+a-b)\;,(a+b-c)>0$
$$\displaystyle \frac{(b+c-a)+(c+a-b)+(a+b-c)}{3}\geq \sqrt[3]{(b+c-a)\cdot(c+a-b)\cdot(a+b-c)}$$
So we get $\displaystyle (a+b+c)\geq 3\sqrt[3]{(b+c-a)\cdot(c+a-b)\cdot(a+b-c)}$
But I did not understand how can I prove above inequality
help Required
Thanks

For a triangle
$\Delta = \frac{abc}{4R} = rs$
Now in your inequality
you can put in the values to get
$R \ge 2r$
This is known to be true since the distance between incentre and circumcentre $d^2 = R(R-2r)$
Thus your inequality is proved  
