For positive real numbers $\Large a, b, c$, find the minimum integer value possible of the following equation:
$\Large 6a^3+9b^3+32c^3+\frac{1}{4abc}$

Question

For positive real numbers $\Large a, b, c$, find the minimum integer value possible of the following equation:
$\Large 6a^3+9b^3+32c^3+\frac{1}{4abc}$

kainui · Answer

Hmm, is that last positive sign supposed to be an equals sign, or is there more here?

aum · Answer

An equation has to have an equal sign which is missing here.

aum · Answer

Is it supposed to be "the following expression" instead of "the following equation" ?

anonymous · Answer

@aum yes
Sorry I write wrong

anonymous · Answer

Hint: Using the AM-GM Inequality

anonymous · Answer

The main key in solving this equation is using the AM-GM Inequality.

-Using the AM-GM Inequality with $\Large x_1=6a^3, x_2=9b^3, x_3=32c^3$ we obtain,

$\Large \frac{6a^3+9b^3+32c^3}{3} \ge \sqrt[3]{1728a^3b^3c^3}$

Which simplifies nicely into,

$\Large 6a^3+9b^3+32c^3 \ge 36abc$

Thus, adding $\Large \frac{1}{4abc}$  on both sides we get,

$\Large 6a^3+9b^3+32c^3+ \frac{1}{4abc} \ge 36abc+\frac{1}{4abc}$

-Applying AM-GM on RHS on the inequality, $\Large x_1=36abc, x_2=\frac{1}{4abc}$, we attain,

$\huge \frac{36abc+ \frac{1}{4abc}}{2}\ge \sqrt[2]{36abc\times \frac{1}{4abc}}\\huge 36abc+\frac{1}{4abc}\ge 6$

Thus,

$\Large 6a^3+9b^3+32c^3+ \frac{1}{4abc} \ge 36abc+\frac{1}{4abc}\ge 6$

Minimum value is 6.

aum · Answer

Nice problem!

anonymous · Answer

:) thanks