Minimum value #4

First, put $x=\frac{1}{a}, y=\frac{1}{b}, z=c$ .

The first equality transforms to $x+y+z=3$ .

The second equation transforms to $\sqrt{\frac{1}{x+y+xy}}+\sqrt{\frac{1}{y+z+yz}}+\sqrt{\frac{1}{z+x+zx}}$ .

Now, by AM - HM, we have $\sqrt{\frac{1}{x+y+xy}}+\sqrt{\frac{1}{y+z+yz}}+\sqrt{\frac{1}{z+x+zx}}\geq \frac{9}{\sqrt{x+y+xy}+\sqrt{y+z+yz}+\sqrt{z+x+zx}}$ .

This reduces the problem to maximizing $\sqrt{x+y+xy}+\sqrt{y+z+yz}+\sqrt{z+x+zx}$ given that $x+y+z=3$ .

By the Cauchy-Schwarz inequality, we have $((\sqrt{x+y+xy})^{2}+(\sqrt{y+z+yz})^{2}+(\sqrt{z+x+zx})^{2})(1^{2}+1^{2}+1^{2})\geq (\sqrt{x+y+xy}+\sqrt{y+z+yz}+\sqrt{z+x+zx})^{2}$

That is, $\sqrt{x+y+xy}+\sqrt{y+z+yz}+\sqrt{z+x+zx}\leq \sqrt{3(2(x+y+z)+xy+yz+zx)}$

$\sqrt{x+y+xy}+\sqrt{y+z+yz}+\sqrt{z+x+zx}\leq \sqrt{3(6+xy+yz+zx)}$

$\sqrt{x+y+xy}+\sqrt{y+z+yz}+\sqrt{z+x+zx}\leq \sqrt{3(6+\frac{9-(x^{2}+y^{2}+z^{2})}{2})}$

By using the Cauchy Schwarz Inequality again (in the same way as the first time we used it), we get $x^{2}+y^{2}+z^{2}\geq \frac{(x+y+z)^{2}}{3}=3$

Thus, $\sqrt{x+y+xy}+\sqrt{y+z+yz}+\sqrt{z+x+zx}\leq 3\sqrt{3}$

Therefore, $\sqrt{\frac{1}{x+y+xy}}+\sqrt{\frac{1}{y+z+yz}}+\sqrt{\frac{1}{z+x+zx}}\geq \sqrt{3}$ .