Problem #2

Algebra Level 4

What is the minimum value of $\large\frac{x^{3}+2y^{3}+4z^{3}}{xyz}$ where $x, y, z$ are positive real numbers?

If you don't know where to start you may read up on well-known inequalities.

This problem is part of the set Easy Problems

The answer is 6.

1 solution

A Former Brilliant Member
Oct 23, 2014

$Since\quad we\quad know\quad that\quad x,y\quad and\quad z\quad ar\quad positive\quad integers,\quad the\quad \\ first\quad thing\quad that\quad must\quad come\quad to\quad our\quad mind\quad is\quad the\quad AM-GM\\ inequality.\\ So,\quad applying\quad the\quad AM\ge GM\quad inequality\quad for\quad { x }^{ 3 },\quad 2{ y }^{ 3 },\quad 4{ z }^{ 3 }\\ \frac { { x }^{ 3 }+2{ y }^{ 3 }+4{ z }^{ 3 } }{ 3 } \ge { ({ x }^{ 3 }.2{ y }^{ 3 }.4{ z }^{ 3 }) }^{ \frac { 1 }{ 3 } }\\ \\ \frac { { x }^{ 3 }+2{ y }^{ 3 }+4{ z }^{ 3 } }{ 3 } \ge { (8{ x }^{ 3 }{ y }^{ 3 }{ z }^{ 3 }) }^{ \frac { 1 }{ 3 } }\\ \\ \frac { { x }^{ 3 }+2{ y }^{ 3 }+4{ z }^{ 3 } }{ 3 } \ge 2xyz\\ \\ \frac { { x }^{ 3 }+2{ y }^{ 3 }+4{ z }^{ 3 } }{ xyz } \ge 6\\ \\ { \frac { { x }^{ 3 }+2{ y }^{ 3 }+4{ z }^{ 3 } }{ xyz } }_{ min. }=\quad 6$

ça ne marche pas toujours.

Omar El Mokhtar - 6 years, 6 months ago

Que voulez-vous dire?

A Former Brilliant Member - 6 years, 5 months ago

:o Est-ce-que tu parle le francais????

Krishna Ar - 6 years, 5 months ago

Problem #2

The answer is 6.

1 solution

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