Inequalities

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High School Math Grades 9-12, Ages 13-18

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High School Math Grades 9-12, Ages 13-18

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lgx57

#1 h Mar 23, 2025, 2:25 PM

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Let $a,b,c>0$ , $\frac{a^2+b^2+c^2}{ab+bc+ca}=2$ , find the minimum of

$\frac{a^3+b^3+c^3}{abc}$

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sqing

#2 h Mar 23, 2025, 2:52 PM

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lgx57 wrote:

Let $a,b,c>0$ , $\frac{a^2+b^2+c^2}{ab+bc+ca}=2$ , find the minimum of

$\frac{a^3+b^3+c^3}{abc}$

https://artofproblemsolving.com/community/c6h2886094p25669333

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#4 h Yesterday at 4:54 AM

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Let $ab + bc + ca = 1$ WLOG so that $a^2 + b^2 + c^2 = 2 \implies a+b+c = 2.$ Then $a^3 + b^3 + c^3 = (a+b+c)(a^2+b^2+c^2-ab-bc-ca) + 3abc = 2 + 3 abc.$ Thus the desired expression is $3 + \frac{2}{abc}$ so we want to maximize $abc.$ So we want the cubic $x^3 - 2x^2 + x - P$ to be tangent to the $x$ -axis at some point where $P = abc.$ Its local minima and maxima are at $x = 1$ and $x = \frac{1}{3}$ , of which $x=\frac{1}{3}$ is the local maximum. Plugging in gives $x^2 - 2x^2 + x - P = \left(x-\frac{1}{3}\right)^2 \left(x-\frac{4}{3}\right) \implies P = \frac{4}{27}.$ Thus the answer is $3 + \frac{54}{4} = \frac{33}{2}.$ Equality at $a=b=\frac{1}{3}$ and $c = \frac{4}{3}$

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