USAMO 2018: Show that $2(ab+bc+ca) + 4 \min(a^2,b^2,c^2) \geq a^2 + b^2 + c^2$ Here is question 1 from USAMO 2018 Q1 (held in April):

Let $a,b,c$ be positive real numbers such that $a+b+c = 4 \sqrt[3]{abc}$. 
   Prove that: $$2(ab+bc+ca) + 4 \min(a^2,b^2,c^2) \geq a^2 + b^2 + c^2$$ 

This question is on symmetric polynomials.  I recall as many facts as I can think of regarding inequalities.  (This test was closed book).


*

*AM-GM inequality: $a + b + c \geq 3 \sqrt[3]{abc}$ so the equality there is strange.

*quadratic mean inequality suggests $3(a^2 + b^2 + c^2) > a + b + c $.

*The $\min(a^2,b^2,c^2)$ on the left side makes things difficult since we can't make it smaller.

*I am still looking for other inequalities that might work.


It's tempting to race through this problem with the first solution that comes to mind.  I'm especially interested in some kind of organizing principle.
 A: Since our inequality and condition are symmetric and homogeneous, 
we can assume that  $abc=1$ and $a\geq b\geq c$.
Thus, $a+b+c=4$ and we need to prove that
$$2(ab+ac+bc)\geq a^2+b^2-3c^2$$ or
$$(a+b+c)^2\geq2(a^2+b^2-c^2)$$ or
$$8\geq a^2+b^2-c^2$$ or $$8\geq a^2+b^2-(4-a-b)^2$$ or
$$8\geq a^2+b^2-(16+a^2+b^2-8a-8b+2ab)$$ or
$$12-4(a+b)+ab\geq0$$ or $$12-4(4-c)+\frac{1}{c}\geq0$$ or
$$(2c-1)^2\geq0.$$
Done!
A: Without loss of generality, let $a\le b=ax\le c=axy, x\ge 1, y\ge 1$.
Then:
$$a+b+c = 4 \sqrt[3]{abc} \Rightarrow a+ax+axy=4\sqrt[3]{a(ax)(axy)} \Rightarrow 1+x+xy=4x^{\frac23}y^{\frac13} \qquad (1)$$
Also:
$$a+b+c = 4 \sqrt[3]{abc} \Rightarrow a^2+b^2+c^2=16\sqrt[3]{(abc)^2}-2(ab+bc+ca) \qquad (2)$$
Plugging $(2)$ and then $(1)$ into the given inequality:
$$2(ab+bc+ca) + 4 \min(a^2,b^2,c^2) \geq a^2 + b^2 + c^2 \overbrace{\Rightarrow}^{(2)}\\
2(ab+bc+ca)+4a^2\ge 16\sqrt[3]{(abc)^2}-2(ab+bc+ca) \Rightarrow \\
(a(ax)+(ax)(axy)+(axy)a)+a^2\ge 4\sqrt[3]{(a(ax)(axy))^2} \Rightarrow\\
x+x^2y+xy+1\ge 4x\sqrt[3]{xy^2} \overbrace{\Rightarrow}^{(1)}\\
4x^{\frac23}y^{\frac13}+x^2y\ge 4x^{\frac43}y^{\frac23} \Rightarrow\\
\left(x^{\frac23}y^{\frac13}\right)^2-4\left(x^{\frac23}y^{\frac13}\right)+4\ge 0 \Rightarrow \\
\left(x^{\frac23}y^{\frac13}-2\right)^2\ge 0.$$
A: $ 2(ab+bc+ca)$+$ 4\min( a^2,b^2,c^2)$=$ 4(ab+bc+ca)+ 4\min(a^2,b^2,c^2)-2(ab+bc+ca)$ 
Now, suppose $a≤b≤c$ without losing generality.
Hence,
$ 2(ab+bc+ca)$+$ 4\min( a^2,b^2,c^2)$
=$ 4(ab+bc+ca)+ 4a^2-2(ab+bc+ca)$
= $4((a+b+c)a+bc)-2(ab+bc+ca)$
=$4(4a(abc)^{1/3})+bc)-2(ab+bc+ca)≥ 16(abc)^{2/3}-2(ab+bc+ca)
=(a+b+c)^2-2(ab+bc+ca)=a^2+b^2+c^2$
