contest math - Inequality with $a,b,cin{}mathbb{R}$.

Prove that for every positive real numbers $a,b$ and $c$ we have

$$(a+b+c)^5\ge 81(a^2+b^2+c^2)abc.$$

I tried using the u,v,w method by substituting

$$a+b+c=3u$$
$$ab+bc+ca=3v^2$$
$$abc=w^3$$

From which it suffices to show $u^5-3u^2w^3+2v^2w^3\geq0$. Im quite stuck here and unable to proceed. Also I know that equality occurs when $a=b=c$.

Answer

EDIT: I tried adding vvnitram's condition, but it doesn't sit well with me, because from cauchy-scharwz $(1 + 1 + 1)(a^2 + b^2 + c^2) \geq (a + b + c)^2$

This inequality should only work for non negative numbers. From AM-GM

$$\begin{align} \frac{a + b + c}{3} &\geq (abc)^{\frac{1}{3}}\\ (a + b + c)^3 &\geq 27abc \end{align}$$
Now for positive numbers $a$, $b$ and $c$ it follows that
$$\begin{align} (a + b + c)^2 = a^2 + b^2 +c^2 + 2ab + 2ac + 2bc \geq a^2 + b^2 + c^2 \end{align}$$
Using vvnitram's condition, assuming $a \geq b \geq c$ then
$$\begin{equation} (a + b + c)^2 \geq 3(a^2 + b^2 + c^2) \end{equation}$$
(because $ab > b^2$)
If $a \geq b$ and $c \geq d$ for positive numbers then $ac \geq bd$. So using the above two inequalities
$$\begin{align} (a + b + c)^3 \cdot (a + b + c)^2 &\geq 27abc \cdot 3(a^2 + b^2 + c^2)\\ (a + b + c)^5 &\geq 81abc(a^2 + b^2 + c^2) \end{align}$$