<s>[INST] Prove the following algebraic asymmetric inequality: $\sqrt{3abc(a + b + c)} + 2(a - c)^2 \geq a^2 + b^2 + c^2$ 
Consider $a \geq b \geq c \geq 0$ real numbers. Prove that:
$$\sqrt{3abc(a + b + c)} + 2(a - c)^2 \geq a^2 + b^2 + c^2$$

Source: RMM 2022
Comments & discussion:


A weaker form of the inequality may be obtained by using Newton's 3 variable inequality:
$$ab + bc + ca \geq \sqrt{3abc(a + b + c)}$$
This means that the new inequality would look like:
$$ab + bc - 3ac + a^2 + c^2 \leq b^2$$
Which will rewrite as:
$$(b - a)(b - c) \leq 0$$
Which is obviously true


However, I am unable to reduce the more tough form to a simillar expression.
 [/INST] Using $ab + bc + ca \ge \sqrt{3abc(a + b + c)}$, we have
$$\sqrt{3abc(a+b+c)}
\ge \frac{3abc(a+b+c)}{ab + bc + ca}.$$
Using $a \ge b \ge c$, we have
$$2(a-c)^2 = (a - c)^2 + (a - b + b - c)^2 \ge (a - c)^2 + (a - b)^2 + (b - c)^2.$$
It suffices to prove that
$$ \frac{3abc(a+b+c)}{ab + bc + ca} + (a - c)^2 + (a - b)^2 + (b - c)^2 \ge a^2 + b^2 + c^2$$
or
$$\frac{3abc(a+b+c)}{ab + bc + ca} + a^2 + b^2 + c^2 - 2(ab + bc + ca)\ge 0. $$
Let $p = a + b + c, q = ab + bc + ca, r = abc$. We need to prove that
$$\frac{3pr}{q} + p^2 - 4q \ge 0. \tag{1}$$
If $p^2 - 4q \ge 0$, (1) is true.
If $p^2 - 4q < 0$, using three degree Schur $r\ge \frac{4pq - p^3}{9}$, we have
$$\frac{3pr}{q} + p^2 - 4q
\ge \frac{3p}{q}\cdot \frac{4pq - p^3}{9} + p^2 - 4q
= \frac{(4q - p^2)(p^2 - 3q)}{3q} \ge 0$$
where we have used $p^2-3q = a^2 + b^2 + c^2 - ab - bc - ca \ge 0$.
We are done.
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