Minimum of $\left(a + b + c + d\right)\left(\frac{1}{a} + \frac{1}{b} + \frac{4}{c} + \frac{16}{d}\right)$ If $a$, $b$, $c$, $d$ are positive integers, find the minimum value of
$$P = \left(a + b + c + d\right)\left(\frac{1}{a} + \frac{1}{b} + \frac{4}{c} + \frac{16}{d}\right)$$
and the values of $a$, $b$, $c$, $d$ when it is reached.
My try:
$$\left.
\begin{array}{l}
a + b + c + d \ge 4\sqrt[4]{abcd}\\
\frac{1}{a} + \frac{1}{b} + \frac{4}{c} + \frac{16}{d} \ge 4\sqrt[4]{\frac{64}{abcd}}
\end{array}
\right\}
\Rightarrow P \ge 32\sqrt{2}$$
I have used mean inequalities, but that doesn't mean that I have found the minimum value. Also, I have found a similar exercise here (exercise #5), but the author shows that $P \ge 64$, which is greater than what have I found.
Can you help me solve the problem, please? Thanks!

If you want to use the AM-GM Inequality, it can be done as follows.  Observe that
$$a+b+c+d=a+b+2\left(\frac{c}{2}\right)+4\left(\frac{d}{4}\right)\geq 8\sqrt[8]{ab\left(\frac{c}{2}\right)^2\left(\frac{d}{4}\right)^4}$$
and that
$$\frac{1}{a}+\frac{1}{b}+\frac{4}{c}+\frac{16}{d}=\frac{1}{a}+\frac{1}{b}+2\left(\frac{2}{c}\right)+4\left(\frac{4}{d}\right)\geq 8\sqrt[8]{\left(\frac1a\right)\left(\frac1b\right)\left(\frac{2}{c}\right)^2\left(\frac{4}{d}\right)^4}\,.$$
However, using the Cauchy-Schwarz Inequality is probably the easiest way.  (The equality holds iff there exists $\lambda >0$ such that $(a,b,c,d)=(\lambda,\lambda,2\lambda,4\lambda)$.)
