Malik wrote: 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 of $a, b$ and $c$ is equality attained? The minimum of $f(x)=ux^3+\frac 1{vx}$ when $u,v,x>0$ is reached when $3uvx^4=1$ *edited * : $v^2\to v $ Considering $a,b$ as parameters and $c$ as variable, minimum is reached when $3\cdot 2^7abc^4=1$ Considering $a,c$ as parameters and $b$ as variable, minimum is reached when $3^3\cdot 2^2ab^4c=1$ Considering $b,c$ as parameters and $a$ as variable, minimum is reached when $3^2\cdot 2^3abc^4=1$ Multiplying these three equalities gives $12abc=1$ and then : $12abc=1$ and $3\cdot 2^7abc^4=1$ gives $c=2^{-\frac 53}$ $12abc=1$ and $3^3\cdot 2^2ab^4c=1$ gives $b=3^{-\frac 23}$ $12abc=1$ and $3^2\cdot 2^3abc^4=1$ gives $a=6^{-\frac 13}$ Hence the answer : $\boxed{(a,b,c)=\left(\frac 1{\sqrt[3]6},\frac 1{\sqrt[3]9},\frac 1{\sqrt[3]{32}}\right)}$ and then value of expression indeed is $6$

$ 6a^{3}+9b^{3}+32c^{3}+\frac{1}{4abc}=6\left(a^3+\frac{(3b)^3}{18}+\frac{(4c)^3}{12}+\frac{1}{2a(3b)(4c)}\right)$ $\ge6\left(3\sqrt[3]{\frac{a^3(3b)^3(4c)^3}{2^33^3}}+\frac{1}{2a(3b)(4c)}\right)=6\left(\frac{a(3b)(4c)}{2}+\frac{1}{2a(3b)(4c)}\right)\ge 6$. The equality hold when $a(3b)(4c)=1$ and $a^3=\frac{(3b)^3}{18}=\frac{(4c)^3}{12}\Leftrightarrow a^3=\frac 16,\;b^3=\frac19,\, c^3=\frac{1}{32}$.

Hey pco! I don't understand where you got this from, can you explain? Considering a,b as parameters and c as variable, minimum is reached when $ 3\cdot 2^{7}abc^{4}=1 $ Thanks!!!

Just consider that we are looking at the minimum of $32x^3+\frac 1{4abx}$ and use the first line of my post with $u=32$ and $v=4ab$

How did this simplest method evaded everybody till now? \[6a^3 + 9b^3 + 32c^3 + \frac{1}{4abc} = 6a^3 + 9b^3 + 32c^3 + \frac{1}{12abc}+ \frac{1}{12abc}+ \frac{1}{12abc}\geq 6\sqrt[6]{\frac {6\cdot 9\cdot 32} {12^3}} = 6,\] with equality for $6a^3 = 9b^3 = 32c^3 = \frac{1}{12abc}$, hence when $abc=\frac{1} {12}$, and so $a=\frac {1} {\sqrt[3]{6}}$, $b=\frac {1} {\sqrt[3]{9}}$, $c=\frac {1} {\sqrt[3]{32}}$.