Hint to an Inequality I have the following problem in my book-

Prove that for all non-negative real numbers $x, y, z$- $$ 6(x+y-z)(x^2+y^2+z^2)+27xyz \le 10(x^2+y^2+z^2)^{3\over 2} $$

And the solution:
$$\color{red}{10(x^2+y^2+z^2)^{3\over 2}-6(x+y-z)(x^2+y^2+z^2)}=(x^2+y^2+z^2)(10\sqrt{x^2+y^2+z^2}-6(x+y-z))$$ $$ =(x^2+y^2+z^2)\left(\color{red}{{10\over 3}\sqrt{(x^2+y^2+z^2)(2^2+2^2+1^2)}}-6(x+y-z)\right) $$ By C-S,
$$ \ge(x^2+y^2+z^2)\left({10(2x+2y+z)\over 3}-6(x+y-z)\right)={(x^2+y^2+z^2)(2x+2y+28z)\over 3}$$By AM-GM,
$ x^2+y^2+z^2 \ge 9\sqrt[9]{x^8y^8z^2\over 4^8} $ and $ 2x+2y+28z\ge 9\sqrt[9]{4^8xyz^7} $
and multiplication accompanied by division with 3 ends the proof.
The validity of the proof is undoubted, but I want to know that how the author comes up with such an idea?
What has the question reflected that hints the major steps to the proof in red?
Note:- I want just the idea that leads to the proof, not a solution.

Because the equality occurs for $(x,y,z)=(2,2,1)$ and we want to delete a radical.
We can try to make it by using C-S:
$$\sqrt{(2^2+2^2+1^2)(x^2+y^2+z^2)}\geq2x+2y+z$$  and it gives a right inequality.
The rest is smooth enough.
