Find the minimum value of $$p=3x+\frac{1}{15x}+5y+\frac{25}{y}+z+\frac{1}{36z},$$ where $x,y,z\in \mathbb{R}^+.$
Applying the AM-GM inequality,
$$
\begin{aligned}\frac{p}{6} & \geqslant\left(3x\cdot\frac{1}{15x}\cdot 5y\cdot\frac{25}{y}\cdot z\cdot \frac{1}{36z}\right)^{1/6} \\
\frac{p}{6} & \geqslant \left(\frac{5}{6}\right)^{1/3}\\
p & \geqslant 6\left(\frac{5}{6}\right)^{1/3}
\end{aligned}$$
$$\implies \text{The minimum value of the expression is } 6\left(\frac{5}{6}\right)^{1/3}
$$
Now, consider $f(x)=3x+\dfrac{1}{15x},\ g(y) = 5y+\dfrac{25}{y}$ and $h(z)=z+\dfrac{1}{36z}.$
$$\begin{aligned}f'(x) &= \frac{\mathrm d}{\mathrm dx}\left(3x+\frac{1}{15x}\right) = 3 – \frac{1}{15x^2}\\f''(x)&=\frac{2}{15x^3}\end{aligned}$$
At the critical points, $f'(x) = 0 \implies x = \dfrac{\pm1}{3\sqrt{5}}.$
$f''\left(\dfrac{1}{3\sqrt5}\right) > 0\implies f(x)$ has a local minima at $x = \dfrac{1}{3\sqrt5}.$
Similarly, the local minima of $g(y)$ is at $y = \sqrt5$ and the local minima of $h(z)$ is at $z=1/6.$
Substituting these values into the original expression, the minimum value of the expression comes out to be
$$
\begin{aligned}
p & = f\left(\dfrac{1}{3\sqrt5}\right)+g\left(\sqrt5\right)+h\left(\frac{1}{6}\right) \\
& = \frac{1}{3} + \frac{52}{\sqrt5}.
\end{aligned}
$$
The answer according to the AM-GM inequality is $\approx5.646$ and according to calculus is $\approx23.588$, which are way off. Also, the functions have only two points of inflection, one is the maxima (for values less than $0$) and the other is the minima (for values greater than $0$). As the question clearly states "for $x,y,z \in \mathbb{R}^+$", the "correct" answer should be $\approx23.588$, shouldn't it?
Why are the answers different? Also, is it possible to figure out the individual $x, y$ and $z$ values for which the expression has the minimum value (in case of AM-GM inequality)?
Note: This question is from a Test and the correct answer according to the "test creators" is $$6\left(\dfrac{5}{6}\right)^{1/3}.$$
$\text{Graph made using Desmos.}$
Best Answer
The three functions $$x\mapsto3x+{1\over15x},\qquad y\mapsto5y+{25\over y},\qquad z\mapsto z+{1\over36z}$$ have similar looking U-graphs when restricting the variables to the positive real axis, having a respective global minimum at $$x_0={1\over 3\sqrt{5}},\qquad y_0=\sqrt{5},\qquad z_0={1\over6}\ .$$ Choosing these values gives $$p(x_0,y_0,z_0)={1\over3}+{52\over\sqrt{5}}=23.5884\ .$$ The $\sigma:=6\left({5\over6}\right)^{1/3}=5.646$ suggestion does not make sense in this problem, for the following reason: The AM/GM inequality says that $${x_1+x_2+\ldots+x_6\over6}\geq\bigl(x_1\cdot x_2\cdot\ldots\cdot x_6\bigr)^{1/6}\ ,$$ whereby the $x_i$ can be chosen arbitrarily in ${\mathbb R}_{>0}$. This would lead to the above $\sigma$. But in the case at hand there are additional constraints, asking for $x_1\cdot x_2={1\over5}$, $\>x_3\cdot x_4=125$, and $x_5\cdot x_6={1\over36}$. This implies that the "overall" AM/GM bound cannot be reached by the admissible $x_i$ of the problem.