Simple AM-GM Application

Algebra Level 4

The minimum value of $ax^3+by^3+cz^2+\frac{d}{xyz}$ for fixed $a$ , $b$ , $c$ , $d > 0$ and $x, y, z \geq 0$ can be written in the form $\dfrac{n\sqrt[n]{a^2b^2c^3d^6}}{l^{9/n}m^{10/n}}$ , where $l$ , $m$ , and $n$ are positive integers. What is $l+m+n$ ?

The answer is 18.

1 solution

Nishant Sharma
Jun 7, 2014

Since the problem says a single application of A.M-G.M. would do the job so we seek such terms so that their product will result in $x^0y^0z^0$ . L.C.M. of $3$ and $2$ is $6$ . Now applying the inequality:

$\displaystyle\frac{\left(\frac{ax^3}{2}+\frac{ax^3}{2}\right)+\left(\frac{by^3}{2}+\frac{by^3}{2}\right)+\left(\frac{cz^2}{3}+\frac{cz^2}{3}+\frac{cz^2}{3}\right)+\left(\frac{d}{6xyz}+\frac{d}{6xyz}+\frac{d}{6xyz}+\frac{d}{6xyz}+\frac{d}{6xyz}+\frac{d}{6xyz}\right)}{13}\geq\left(\frac{a^2b^2c^3d^6}{2^{10}3^9}\right)^\frac{1}{13}$ . So we get $l+m+n=13+2+3=\boxed{18}$ .

You are a man of genius... I solved it hard.

Myung Chul Lee - 6 years, 12 months ago

Would you mind posting it. Solution might be hard here but it may be easier elsewhere. Who knows?

Matt O - 4 years, 7 months ago

Same. Nice problem . $:)$

Keshav Tiwari - 6 years ago

easy isn't it @Nishant Sharma

Madhukar Thalore - 6 years ago

Simple AM-GM Application

The answer is 18.

1 solution

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