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If $a,b,c$ are real positive numbers. How to find the minimum for:

$$6a^3+9b^3+32c^3+\frac{1}{4abc}$$

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2 Answers 2

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Using the Cauchy inequality with 6 positive numbers:

$6a^3$

$9b^3$

$32c^3$

and $3$ numbers $\displaystyle\frac{1}{12abc}.$

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How do you find a minimum? Take a derivative (yes, there are other ways). Since you have three variables, you can take a derivative with respect to each and set them to zero. $$18a^2-\frac 1{4a^2bc}=0\\27b^2-\frac 1{4ab^2c}=0\\96c^2-\frac 1{4abc^2}=0$$ This looks like a real mess, but move the fractions to the other side and mutiply them all $$2^{6}3^6a^2b^2c^2=\frac 1{2^6a^4b^4c^4}\\\frac 1{abc}=12$$ and we can plug this into each of the first three $$18a^2=\frac {3}{a},27b^2=\frac 3b,96c^2=\frac 3c\\a=\frac 1{\sqrt[3]6}, b=\frac 1{\sqrt[3]9},c=\frac 1{\sqrt[3]{32}}$$

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