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 !
 A: Another way.
Since $$\prod_{cyc}(a-1)^2=\prod_{cyc}((b-1)(c-1))\geq0,$$ we can assume $(b-1)(c-1)\geq0$, which gives $$a(b-1)(c-1)\geq0$$ or$$abc\geq ab+ac-a$$ and since $$b^2+c^2\geq\frac{1}{2}(b+c)^2,$$ it's enough to prove that
$$a^2+\frac{1}{2}(b+c)^2+\sqrt2(ab+ac-a)+2\sqrt2+3\geq(2+\sqrt2)(a+b+c)$$ or $$(\sqrt2a+b+c-2-\sqrt2)^2\geq0$$ and we are done!
A: 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.
A: $uvw$ helps. See here: https://math.stackexchange.com/edit-tag-wiki/5758
Let $a+b+c=3u$, $ab+ac+bc=3v^2$ and $abc=w^3$.
Thus, our inequality it's $$9u^2-6v^2+\sqrt2w^3+3+2\sqrt2\geq3(2+\sqrt2)u,$$ which is a linear inequality of $v^2$.
Thus, by $uvw$ it's enough to prove our inequality for equality case of two variables.
Let $b=a$.
Thus, we need to prove a quadratic inequality of $c$ and $\Delta\leq0$ it's an inequality of one variable  $a$.
Can you end it now?
A: 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.
