Let x, y, z be real positive numbers, which are generated from the Ravi transformation of a triangle. Use AM-HM or any other inequality to prove that $$\sqrt{x^2 + y (2 x + 2 z) + 2 y^2 + z^2}<2x+2y$$
The Ravi transformation of a triangle is shown by the following illustration:
Take each side and form it as $D=x+y$, $E=y+x$ and $F=z+x$.
My take: $$\sqrt{x^2 + y (2 x + 2 z) + 2 y^2 + z^2}<2x+2y$$
$$x^2 + y (2 x + 2 z) + 2 y^2 + z^2<4(x+y)^2$$
$$-3 x^2 - 6 x y - 2 y^2 + 2 y z + z^2<0$$
Reciprocal:
$$\frac1{z^2-3 x^2 - 2 y^2 + 2 y z - 6 x y}>0$$
Assuming, $z^2-3 x^2 - 2 y^2\leq0$, then we need only to prove that $$z^2-3 x^2 - 2 y^2<2 y z - 6 x y$$
Hence, $$\frac{z^2-3 x^2 - 2 y^2}{y}<0<2z-6y$$
So if we can prove that $z^2-3 x^2 - 2 y^2\leq0$ holds, then the original inequality holds too.
However how can this be shown ?