# Prove that if $a,b,c$ are positive real numbers

Prove that if $a,b,c$ are positive real numbers, then the least possible value of $$6a^{3}+9b^{3}+32c^{3}+\frac{1}{4abc}$$ is 6. For which values $a,b$ and $c$ is equality attained ?

I know how to prove it, it's just for sharing a new ideas, thanks :)

• Should that be $9b^3$ instead of $9a^3$? – Cataline Aug 9 '15 at 1:07
• @Cataline Yep, thanks – Oiue Aug 9 '15 at 1:08

Outline: Our expression is equal to $$6a^3+9b^3+32c^3+\frac{1}{12abc}+\frac{1}{12abc}+\frac{1}{12abc}.$$ Now use AM-GM. Note that the product of the $6$ terms is equal to $1$.

• Great, and I was hoping for a full solution, thanks, (1+) – Oiue Aug 9 '15 at 1:16
• I guess it would be worthwhile to write out a full solution for those who don't know the meaning of AM-GM. But I imagine you do, – André Nicolas Aug 9 '15 at 1:26
• yes I know, and this is information to the people who don't know what AM-GM means, thanks en.wikipedia.org/wiki/… – Oiue Aug 9 '15 at 1:29
• Maybe I will add detail for future viewers, tomorrow. A "contest kid" can knock it out of the park. – André Nicolas Aug 9 '15 at 2:36

Just for completing @André Nicolas solution:

$$6a^{3}+9b^{3}+32c^{3}+\frac{1}{4abc}=6a^{3}+9b^{3}+32c^{3}+\frac{1}{12abc}+\frac{1}{12abc}+\frac{1}{12abc}\geq 6\sqrt[6]{\frac{6\cdot 9\cdot 32}{12^3}}=6.$$

And I hope to see more solutions, thanks

• With equality precisely if $6a^3=9b^3=32c^3=\frac{1}{12abc}$, which can be solved explicitly. – André Nicolas Aug 9 '15 at 2:38
• @AndréNicolas yes you're right – Oiue Aug 9 '15 at 2:42

Here's another, admittedly less slick, way:

Set $f(a,b,c)=6a^{3}+9b^{3}+32c^{3}+\frac{1}{4abc}$. Then

$f_{a}(a,b,c)=18a^{2}-\frac{1}{4a^{2}bc}$

$f_{b}(a,b,c)=27b^{2}-\frac{1}{4ab^{2}c}$

$f_{c}(a,b,c)=96c^{2}-\frac{1}{4abc^{2}}$.

Then the only critical value $\vec x$ of $f$ for which $a,b,c>0$ is

$(6^{-1/3},3^{-2/3},2^{-5/3})$.

It is easy (by the second derivative test) to check it gives a minimum and that

$f((6^{-1/3},3^{-2/3},2^{-5/3}) )=6$