Prove the inequality $(x+y+z)^3>27(y+z-x)(z+x-y)(x+y-z)$ [duplicate]

The question is to prove $$ (x+y+z)^3>27(-x+y+z)(x-y+z)(x+y-z), $$ for $x\neq y \neq z,\ x,y,z>0$.

My proof below: $$ 27(-x+y+z)(x-y+z)(x+y-z)=-27(x^3+y^3+z^3+2xyz)+27\left\{x^2(y+z)+y^2(z+x)+z^2(x+y)\right\} $$

Adding and subtracting $(x+y+z)^3$ to this, we get the RHS as $$ (x+y+z)^3-28(x^3+y^3+z^3)-60xyz+24\left\{x^2(y+z)+y^2(z+x)+z^2(x+y)\right\} \label{eq1} \tag{1} $$

It can easily be proved that $$ 2(x^3+y^3+z^3)>x^2(y+z)+y^2(z+x)+z^2(x+y)>6xyz $$

This can be converted into equalities as: $$ \begin{aligned} &2(x^3+y^3+z^3)=x^2(y+z)+y^2(z+x)+z^2(x+y)+q_1 \\ &2(x^3+y^3+z^3)=6xyz+q_2 \\ &\{q_1,q_2|\text{ }q_2>q_1>0\} \end{aligned} $$

Using these results in $\eqref{eq1}$, we get $$ \begin{aligned} 27(-x+y+z)(x-y+z)(x+y-z)&=(x+y+z)^3-28(x^3+y^3+z^3)-20(x^3+y^3+z^3)-10q_2+48(x^3+y^3+z^3)-24q_1\\ &=(x+y+z)^3-10q_2-24q_1 \end{aligned} $$

Therefore, we can say $$ (x+y+z)^3>27(-x+y+z)(x-y+z)(x+y-z) $$

My question is: Is there a simpler way to do this?

Let $\ \ \ p = -x+y+z$; $\ \ \ q= x-y+z$; $\ \ \ r = x+y-z$.

Notice that $p+q+r = x+y+z$

In case, $p,q,r>0$.

So, by AM-GM

$$\cfrac{p+q+r}{3} \ge \sqrt[3]{pqr}$$

$$\cfrac{x+y+z}{3} \ge \sqrt[3]{(-x+y+z)(x-y+z)(x+y-z)}$$

On cubing,

$$\cfrac{(x+y+z)^3}{27} \ge (-x+y+z)(x-y+z)(x+y-z) $$

Therefore:

$$\boxed{(x+y+z)^3 \ge 27(-x+y+z)(x-y+z)(x+y-z)}$$

Equality holds for $x=y=z$, but since that doesn't match what the question requires, you can ignore the equality sign.

As fleablood has suggested, note that:

If $x,y,z$ are positive then at most one of the $p,q,r$ is non-positive (as only one of $x,y,z$ can be larger than the sum of the other two) and if that were the case we'd have the simple case that $pqr≤0$ but $(x+y+z)^3>0$.