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.
 
Here is a little puzzle from the book 100 Geometric Games by Pierre Berloquin. The side of a small square is one meter long and the side of a larger square one and a half meters long. One vertex of the large square is at the center of the small square. The side of the large square cuts two sides of the small square into one- third parts and two-thirds parts. What is the area where the squares overlap?

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