Fun with Cyclic

Without loss of generality,one may assume $a+b+c=1$ .

So, now we have to find the minimum value of $f(a,b,c)=\sum_{cyc}\frac{1+\sqrt{a}}{1-a}$ .

By AM-HM inequality, $\frac{\sum_{cyc}(1-a)}{3}≥\frac{3}{\sum_{cyc}\frac{1}{1-a}}$

or, $\sum_{cyc}\frac{1}{1-a}≥\frac{9}{\sum_{cyc}(1-a)}=\frac{9}{3-(a+b+c)}=\frac{9}{2}$

Now let $\sqrt{a}=x,\sqrt{b}=y,\sqrt{c}=z$ . Hence, $x^2+y^2+z^2=1$

Let $g(x,y)=\sum_{cyc}\frac{x}{1-x^2}$ .Now we have to minimize this.

$(y+z)^2≥2(y^2+z^2)=2(1-x^2)$

or, $\frac{x}{1-x^2}≥2\frac{x}{(y+z)^2}$

So, $(x+y+z)g(x,y)≥2(x+y+z)(\sum_{cyc}\frac{x}{(y+z)^2})$ $≥2(\sum_{cyc}\frac{x}{y+z})^2$ $≥2(\frac{3}{2})^2=\frac{9}{2}$ (By Nesbitt's Inequality)

By AM-RMS inequality,

$x+y+z≤3\sqrt{\frac{x^2+y^2+z^2}{3}}=\frac{3}{\sqrt{3}}$

or, $\frac{1}{x+y+z}≥\frac{\sqrt{3}}{3}$

Hence, $g(x.y)≥\frac{3\sqrt{3}}{2}$

Therefore, $f(a,b,c)≥\frac{9+3\sqrt{3}}{2}=7.098$