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) $

Question

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 !

Answer 1

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:

1. $p \ge \frac{9\sqrt2}{4}$:

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.

2. $p < \frac{9\sqrt2}{4}$:

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.

Answer 2

$uvw$ 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?

Answer 3

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!

Answer 4

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.