AM-GM inequality for sum of 3 square roots

[I like Serena]

Let $a,b,c$ be positive real numbers with sum $3$.
Prove that $√a+√b+√c≥ab+bc+ca$.

 

[Mathwebster]

Here is my proof

Spoiler

Since $a, b$ and $c$ are positive we can let $a \to a^2, b \to b^2$ and $c \to c^2$ so we want to prove that

$a + b + c \ge a^2b^2 + b^2c^2 + a^2c^2$ if $a^2+b^2+c^2 = 3$

Consider $f(x) = x^4 -3x^2+2x$. It's fairly easy to show that $f(x) \ge 0$ if $x \ge 0.$

Thus, $f(a)+f(b)+f(c) \ge 0$ or

$a^4+b^4+c^4 - 3(a^2+b^2+c^2) + 2(a+b+c) \ge 0$ or re-writing

$a^4+b^4+c^4 + 2(a+b+c) \ge 3(a^2+b^2+c^2)$

so

$a^4+b^4+c^4 + 2(a+b+c) \ge (a^2+b^2+c^2)^2$ since $a^2+b^2+c^2 = 3$.

Expanding gives the desired result.