MHB AM-GM inequality for sum of 3 square roots

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The discussion centers on proving the inequality √a + √b + √c ≥ ab + bc + ca for positive real numbers a, b, and c that sum to 3. The proof begins by applying the AM-GM inequality, which states that the arithmetic mean of non-negative numbers is greater than or equal to their geometric mean. Participants explore various approaches to demonstrate the validity of the inequality, emphasizing the conditions required for equality. The conversation highlights the significance of the inequality in mathematical analysis and its applications. The proof ultimately reinforces the relationship between the sums of square roots and products of the variables.
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Let $a,b,c$ be positive real numbers with sum $3$.
Prove that $√a+√b+√c≥ab+bc+ca$.
 
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Here is my proof
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.
 
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