Find the Minimum

Let us use Titu's Lemma or Cauchy-Schwartz Inequality in Engel form.

$\frac{1}{x}+\frac{1}{y}+\frac{4}{z}+\frac{16}{a}$

Minimum value= $\frac{(\sqrt{1}+\sqrt{1}+\sqrt{4}+\sqrt{16})^{2}}{x+y+z+a}$

which is equal to $\frac{64}{x+y+z+a}$ .

But $x+y+z+a=1$ . So, putting it in the above equation, we get the minimum value

as $\huge{64}$ .

Let $\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{2}{z}+\dfrac{2}{z}+\dfrac{4}{a}+\dfrac{4}{a}+\dfrac{4}{a}+\dfrac{4}{a}=S$

By $AM \geq HM$

$\dfrac{x+y+\dfrac{z}{2}+\dfrac{z}{2}+\dfrac{a}{4}+\dfrac{a}{4}+\dfrac{a}{4}+\dfrac{a}{4}}{8}\geq\dfrac{8}{\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{2}{z}+\dfrac{2}{z}+\dfrac{4}{a}+\dfrac{4}{a}+\dfrac{4}{a}+\dfrac{4}{a}}$

$\dfrac{S}{8}\geq 8$

By this

$\boxed{S \geq 64}$

with positive real numbers u and v we have:

$\dfrac {1}{u}+\dfrac {1}{v} \geq \dfrac {4}{u+v}$

so:

$\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{4}{z}+\dfrac{16}{a}\geq \dfrac{4}{x+y}+\dfrac{4}{z}+\dfrac{16}{a}\geq \dfrac{16}{x+y+z}+\dfrac{16}{a}\geq \dfrac{64}{x+y+z+a}=64$

The minimum occurs when $x=y ; x+y=z ; x+y+z=a ; x+y+z+a=1$ , which means $x=y=\dfrac{1}{8}, z=\dfrac{1}{4}, a=\dfrac{1}{2}$

Answer: 64

One can let g(x,y,z,a) = x + y + z + a - 1 and let f(x,y,z,a) = 1/x + 1/y + 4/z + 16/a and then Use Lagrange Multipliers.