Inspired by nam le - Part 2

$f(x,y,z)=(\sqrt{16x+9}+\sqrt{16y+9}+\sqrt{16z+9})^2$

To achieve an extremum, we either have $\nabla f(x,y,z)=0$ , or we have $x,y,z$ on the boundary of the domain. The domain is $x,y,z\geq-\frac{9}{16}$ *.

Setting $\nabla f=0$ we find $x=y=z=\frac{1}{3}$ is an extremum, which turns out to be the maximum (equal to $129$ ).

If we have just one variable on the boundary, we get a two-variable function $f(x,y)=(\sqrt{16x+9}+\sqrt{16y+9})^2$ with a constraint $x+y=\frac{25}{16}$ , whose local extrema are neither minimums nor maximums.

If we set two of them, i.e. $x=y=-\frac{9}{16}$ , $z=\frac{34}{16}$ we find the minimum $f(-\frac{9}{16},-\frac{9}{16},\frac{34}{16})=43$ .

It's impossible for all three variables to be on the boundary, hence the answer is $43$ .

*Note: Usually for the solution to be complete, we need to consider values of $x,y,z<-\frac{9}{16}$ which yield a real value of $f$ . However in this case there's no such possibility, since for the square of a complex number to be real, it has to take either the form $a\cdot i$ or $b$ . This means that $\sqrt{16x+9}$ , $\sqrt{16y+9}$ , $\sqrt{16z+9}$ are either all real or all non-real, but for them to all be non-real, $x,y,z\leq-\frac{9}{16}$ , which is incompatible with the restraint $x+y+z=1$ .