No Additional Condition Needed

The partial derivatives are zero when $a^2+1=a(a+b+c+1)$ , $b^2+1=b(a+b+c+1)$ and $c^2+1=c(a+b+c+1)$ . Equivalently, it is required that $a+\frac{1}{a}=b+\frac{1}{b}=c+\frac{1}{c}=a+b+c+1$ Up to a permutation of the variables, we have either $a=b=c$ or $a=b=\frac{1}{c}$ . The second case implies $a=-1$ , which we reject. In the first case we find that $a=b=c=\frac{1}{2}$ (again, we reject $a=-1$ ) and the maximum value will be $\frac{16}{5}=\boxed{3.2}$ .

I did It like this. Don't know if its flawed. tell me- let a=tan x, b=tan y,c=tan z so x,y,z>0, all taking only principle values. therefore, LHS= $( tan x + tan y + tan z + 1)^{2})/((sec x)^{2}(sec y)^{2}(sec z)^{2})$

= $((tan x +tan y +tan z +1) cos x\times cos y\times cos z)^{2}$

= $(sin x\times cosy\times cosz+cosx \times siny\times cosz+cosx\times cosy\times sinz+cosx\times cos y\times cosz)^{2}$

= $(sin(x+y+z) + sinx \times siny \times sinz + cosx \times cosy \times cosz)^{2}$

Conditional on $x + y + z = \theta$ a constant, because the second derivative of $\ln \sin x$ and $\ln \cos x$ are negative, it tells us that $\prod \sin x$ and $\prod \cos x$ are maximized when $x = y = z$ . Hence, we want to find the maximum of $\sin 3x + \sin ^3 x + \cos ^3 x$

It remains to verify (my approach was to use calculus) that the maximum occurs at $x = 2 \pi - 2 \tan^{-1} ( 2 - \sqrt{5} )$ , and has the value of $\frac{4}{ \sqrt{5} }$ . Thus, the max of the expression is $\frac{16}{5}$ .

Suppose a = b = c. Then $f (a) = (3a+1)^{2}/(a^{2}+1)^{3}$ . And $d (f (a))/(da) = [6(3a+1)(a^{2}+1)^{3}-6(3a+1)^{2} \times (a^{2}+1)^{2} \times a]/(a^{2}+1)^{6}$ $= 6 [(3a+1)(a^{2}+1) - (3a+1)^{2} \times a]/(a^{2}+1)^{4}$ . When we make $d (f (a))/(da) = 0$ then we get $a^2+1 -(3a+1) \times a = 0$ $2a^{2}+a-1=0$ So a=1/2. Finally, $f (a=1/2)=16/5$ .