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

Thanks mathlove,phira,and user for solution.
My prove for $(b+c-a)\cdot(c+a-b)\cdot(a+b-c)\leq abc$, where $a,b,c$ are the sides of a $\triangle$.
Using $\;\;\;\; \{b+(c-a)\}\cdot \{b-(c-a)\} = b^2-(c-a)^2\leq b^2$ 
similarly $ \{c+(a-b)\}\cdot \{c-(a-b)\} = c^2-(a-b)^2\leq c^2$
similarly $\{a+(b-c)\}\cdot \{a-(b-c)\} = a^2-(b-c)^2\leq a^2$
Multimply these three equations, we get $(b+c-a)\cdot(c+a-b)\cdot(a+b-c)\leq abc$
and equality hold when $a=b=c$
