Find the minimum value of $f(a,b,c)$ Given that $a,b,c>0$, find the minimum value of the function $$f(a,b,c)=\frac{a^3}{4b}+\frac{b}{8c^3}+\frac{1+c}{2a}$$ Need some hints. It probably requires AM-GM inequality, I am just unable to see the terms... Thanks.
 A: AM-GM is messy in this case. Let $\alpha \in (0, \frac35)$. Then we have
$$f(a, b, c) = \frac15\cdot\frac{a^3}{\frac45 b}+\frac15\cdot\frac{b}{\frac85c^3}+\alpha\cdot \frac1{2\alpha a}+(\frac35-\alpha)\cdot\frac{c}{2(\frac35-\alpha)a}\ge \frac{5^{1-\alpha } \left(\frac{1}{3-5 \alpha }\right)^{\frac{3}{5}-\alpha } \left(\frac{1}{\alpha }\right)^{\alpha } c^{-\alpha }}{2\cdot 2^{3/5}}$$
Now for any $\alpha$ in the interval, we can find $a, b, c$ to give equality, and such minimum keeps decreasing with $\alpha$, so there is really no minimum but only an infimum.  In the limit as $\alpha \to 0^+$, we get it as $f(a, b, c) > \dfrac5{2\cdot 6^{3/5}} \approx 0.8532$.
A: In a first step keep $a$ and $c$ fixed, and minimize
$$g(a,b,c):={a^3\over 4b}+{b\over 8c^3}$$
by choosing $b$ properly. The optimal choice is $b=\sqrt{2a^3c^3}$, and we obtain
$$g_\min(a,c)={1\over4}\sqrt{{2a^3\over c^3}}\ .$$
Therefore we are left with the problem to minimize
$$h(a,c):={1\over4}\sqrt{{2a^3\over c^3}}+{1+c\over2a}$$
over positive $a$ and $c$. Determine the optimal $a$ for given $c$ and then $h_\min(c)$ as a function of the single variable $c$. Finally minimize this by a proper choice of $c$.
