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: $$\left. \begin{array}{l} a + b + c + d \ge 4\sqrt{abcd}\\ \frac{1}{a} + \frac{1}{b} + \frac{4}{c} + \frac{16}{d} \ge 4\sqrt{\frac{64}{abcd}} \end{array} \right\} \Rightarrow P \ge 32\sqrt{2}$$

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!

• Hint: Use CSB inequality. (As pointed out in the hyperlinked article.) Jul 19 '18 at 10:29
• @JoseArnaldoBebitaDris, I have found a better value than the one with CSB. Jul 19 '18 at 10:33
• @IulianOleniuc Your lower bound is not sharp. So, it is not really better. Jul 19 '18 at 10:39

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