Let $a,b,c$ be non-negative real numbers .Prove that : $ a^2+b^2+c^2 +\sqrt{2} abc + 2\sqrt{2} +3 \geq (2+\sqrt {2} )(a+b+c) $ Let $a,b,c$ be non-negative real numbers
Prove that : $ a^2+b^2+c^2 +\sqrt{2} abc + 2\sqrt{2} +3 \geq (2+\sqrt {2} )(a+b+c) $
My idea is to use the $(p,q,r)$ method:
$p=a+b+c$
$q=ab+bc+ca$
$r = abc $
$\Rightarrow a^2+b^2+c^2 = p^2-2q $
$ a^2+b^2+c^2 +\sqrt{2} abc + 2\sqrt{2} +3 \geq (2+\sqrt {2} )(a+b+c) $
or
$ p^2-2q +\sqrt{2} r + 2\sqrt{2} +3 \geq (2+\sqrt {2} )p $
or
$ (p - (1+\sqrt{2} ))^2 -2q +\sqrt{2} r \geq 0 $
or
The problem I am facing is exactly what I want to prove : $\sqrt{2} r \geq 2q$ is completely wrong .
I hope to get help from everyone. Thanks very much !

Alternative proof:
The desired inequality is easily written as
$$\left(a + \frac{\sqrt2\, bc - 2 - \sqrt2}{2}\right)^2 + \frac{2 - c^2}{2}\left(b - \frac{2 + \sqrt2 - c - c\sqrt2}{2 - c^2}\right)^2 + \frac{c(c - 1)^2}{c + \sqrt2} \ge 0.$$
So, if $c < \sqrt2$, the desired inequality is true.
If $c \ge \sqrt2$, it suffices to prove that
$$a^2 + b^2 + c^2 + \sqrt2\, ab \cdot \sqrt2 + 2\sqrt2 + 3 \ge (2 + \sqrt2)(a + b + c)$$
or
$$\frac14(2c - 2 - \sqrt2)^2 + \frac14(2a + 2b - 2 - \sqrt2)^2 \ge 0$$
which is true.
We are done.
