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- Mar 5, 2012

- 9,001

Let $a,b,c$ be positive real numbers with sum $3$.

Prove that $√a+√b+√c≥ab+bc+ca$.

Prove that $√a+√b+√c≥ab+bc+ca$.

- Thread starter Klaas van Aarsen
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- Thread starter
- Admin
- #1

- Mar 5, 2012

- 9,001

Let $a,b,c$ be positive real numbers with sum $3$.

Prove that $√a+√b+√c≥ab+bc+ca$.

Prove that $√a+√b+√c≥ab+bc+ca$.

- Jan 26, 2012

- 183

$a + b + c \ge a^2b^2 + b^2c^2 + a^2c^2$ 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.