If $\triangle$ is the area of triangle with side lengths $a,b,c$, then show that $\triangle \le\dfrac{1}{4}\cdot\sqrt{(a+b+c)abc}$. Also show that equality occurs in the above inequality when $a=b=c$
My attempt is as follows:-
$$\triangle=\sqrt{s(s-a)(s-b)(s-c)}$$
$$\triangle=\dfrac{1}{4}\sqrt{(a+b+c)(b+c-a)(a+c-b)(a+b-c)}$$
$$b+c-a>0,a+c-b>0,a+b-c>0$$
So we can apply $A.M\ge G.M$
$$\dfrac{b+c-a+a+c-b+a+b-c}{3}\ge\left((b+c-a)(a+c-b)(a+b-c)\right)^\frac{1}{3}$$ $$\dfrac{a+b+c}{3}\ge\left((b+c-a)(a+c-b)(a+b-c)\right)^\frac{1}{3}\tag{1}$$
As $a>0,b>0,c>0$, we can also say
$$\dfrac{a+b+c}{3}\ge(abc)^\frac{1}{3}\tag{2}$$
But we can't say from this that $(abc)^\frac{1}{3}\ge\left((b+c-a)(a+c-b)(a+b-c)\right)^\frac{1}{3}$
So I tried $G.M\ge H.M$ for $a,b,c$
$$(abc)^\frac{1}{3}\ge\dfrac{3abc}{ab+bc+ca}$$ $$\dfrac{ab+bc+ca}{3}\ge(abc)^\frac{2}{3}\tag{3}$$
Seems like it didn't produce anything useful, so I tried $G.M\ge H.M$ for $b+c-a,a+c-b,a+b-c$
$$\left((b+c-a)(a+c-b)(a+b-c)\right)^\frac{1}{3}\ge\dfrac{3(b+c-a)(a+c-b)(a+b-c)}{((b+c-a)(a+c-b)+(a+c-b)(a+b-c)+(b+c-a)(a+b-c))}$$
$$\left((b+c-a)(a+c-b)(a+b-c)\right)^\frac{1}{3}\ge\dfrac{3(b+c-a)(a+c-b)(a+b-c)}{2ab+2bc+2ca-a^2-b^2-c^2}\tag{4}$$
But again it didn't produce anything useful, so I tried something else.
Applying $A.M\ge G.M$ for $a+b+c,a+b-c,b+c-a,a+c-b$
$$\dfrac{2(a+b+c)}{4}\ge\left((a+b+c)(b+c-a)(a+c-b)(a+b-c)\right)^\frac{1}{4}$$ $$\dfrac{a+b+c}{2}\ge\left((a+b+c)(b+c-a)(a+c-b)(a+b-c)\right)^\frac{1}{4}$$
Squaring both sides
$$\dfrac{(a+b+c)^2}{4}\ge\left((a+b+c)(b+c-a)(a+c-b)(a+b-c)\right)^\frac{1}{2}$$
By cachy's inequality:-
$$(1^2+1^2+1^2)(a^2+b^2+c^2)\ge(a+b+c)^2$$ $$3(a^2+b^2+c^2)\ge(a+b+c)^2$$ $$\dfrac{3(a^2+b^2+c^2)}{4}\ge\dfrac{(a+b+c)^2}{4}\tag{5}$$
Hence
$$\dfrac{3(a^2+b^2+c^2)}{4}\ge\left((a+b+c)(b+c-a)(a+c-b)(a+b-c)\right)^\frac{1}{2}\tag{6}$$
Dividing by $4$
$$\dfrac{3(a^2+b^2+c^2)}{16}\ge\triangle$$
So I tried all these approaches, but unfortunately didn't arrive at the required result. What can we do here? Please help me in this.
