<s>[INST] 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 !
 [/INST] Proceeding along the OP's idea (pqr method):
First of all, we have
\begin{align*}
 p^2 &\ge 3q, \tag{1}\\
 p^3 - 4pq + 9r &\ge 0. \tag{2}
\end{align*}
Note: (2) is just Schur's inequality $a(a - b)(a - c) + b(b - c)(b - a) + c(c - a)(c - b)\ge 0$.
If $q \le \frac{p^2}{4}$, it suffices to prove that
$$p^2 - 2 \cdot \frac{p^2}{4} + 2\sqrt2 + 3 \ge (2 + \sqrt2)p$$
or
$$\frac12(p - 2 - \sqrt2)^2 \ge 0$$
which is true.
If $q > \frac{p^2}{4}$, using (2), it suffices to prove that
$$p^2 - 2q + \sqrt2 \cdot \frac{4pq - p^3}{9}
+ 2\sqrt2 + 3 \ge (2 + \sqrt2)p$$
or
$$\left(\frac{4\sqrt2}{9}p - 2\right)q - \frac{\sqrt2}{9}p^3  + p^2 + 2\sqrt2 + 3 \ge (2 + \sqrt2)p. $$
We split into two cases:

*

*$p \ge \frac{9\sqrt2}{4}$:

It suffices to prove that
$$\left(\frac{4\sqrt2}{9}p - 2\right)\cdot \frac{p^2}{4} - \frac{\sqrt2}{9}p^3  + p^2 + 2\sqrt2 + 3 \ge (2 + \sqrt2)p$$
or
$$\frac12(p - 2 - \sqrt2)^2 \ge 0$$
which is true.


*$p < \frac{9\sqrt2}{4}$:

Using (1), it suffices to prove that
$$\left(\frac{4\sqrt2}{9}p - 2\right)\cdot \frac{p^2}{3} - \frac{\sqrt2}{9}p^3  + p^2 + 2\sqrt2 + 3 \ge (2 + \sqrt2)p $$
or
$$\frac{\sqrt2}{54}(2p + 12 + 9\sqrt2)(p - 3)^2 \ge 0$$
which is true.
We are done.
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