Let: $a,b,c \ge 0$. Prove that: $$(a+bc)^{2}+(b+ca)^{2}+(c+ab)^{2} \ge \sqrt{2}(a+b)(b+c)(c+a)$$
I have a proof here:
WLOG, in 3 numbers $a,b,c$ there will be 2 two numbers that both bigger or smaller than 1 so we can assume 2 numbers are $a$ and $b$.
$$\rightarrow (a-1)(b-1) \ge 0 \Leftrightarrow abc+1\ge ac+bc (*)$$
Also we have: $$(a+bc)^{2} + (b+ca)^{2} \ge \frac{(a+bc+b+ca)^{2}}{2}(**)$$
Now by AM-GM inequality we have:
$$\frac{(a+bc+b+ca)^{2}}{2} + (c+ab)^{2} \ge \sqrt{2}(a+b)(c+1)(c+ab)(***)$$
So from $(**)$ and $(***)$ we have:
$$LHS \ge \sqrt{2}(a+b)(c+1)(c+ab)$$
Then we have to prove:
$$(c+1)(c+ab) \ge (b+c)(c+a)(****)$$
But it is true because $(****)$ can be written as:
$$c(ab+1) \ge c(a+b)$$
So the inequality is proved!
Are there any solution using mixing variable method, because I realize that some inequalities which use $(a-1),(b-1),(c-1)$ will have a mixing variable proof!
Thank a lots for your help!
