What is the maximum value of the function
$$f(a, b, c, d, e, f) = \frac{\sum_{n = 0}^{\infty} (n \cdot [\left(a + b + c \right)^{2} + (d + e + f)^{2}]^{n}}{\sum_{n=0}^{\infty} [(a + b + c)^{2} + (d + e + f)^{2}]^{n}}$$
subject to the constraint
$a + b + c + d + e + f = 1$?
I also know $0 < (a + b + c)^{2} + (d + e + f)^{2} < 1$
Letting $u = (a + b + c)$ and $v = (d + e + f)$, we have
$$f(u, v) = \frac{\sum_{n = 0}^{\infty} n \cdot \left( u^2 + v^2 \right)^{n}}{\sum_{n = 0}^{\infty} (u^2 + v^2)^{n}} = (1 - (u^2 + v^2)) \cdot \sum_{n = 0}^{\infty} n\cdot (u^2 + v^2)^{n}$$
Since $u + v = 1$, we have $u^2 + v^2 = u^2 + (1 - u)^2$. So,
$$f(u, v) = (1 - (u^2 + (1 - u)^2) \cdot \sum_{n = 0}^{\infty} n \cdot \left(u^2 + (1 - u)^2\right)^{n}.$$
$$=(1 - (2u^2 - 2u + 1)) \cdot \sum_{n = 0}^{\infty} n \cdot \left(2u^2 - 2u + 1 \right)$$
$$= (2u - 2u^2) \sum_{n = 0}^{\infty}n \cdot (2u^2 - 2u + 1).$$
But since $\sum_{n = 0}^{\infty} x^{n} = \frac{1}{1 - x}$ for $|x| < 1$, if we take the derivative, we get $\sum_{n = 1}^{\infty} nx^{n - 1} = \frac{1}{(x - 1)^{2}},$ so that
$$f(u, v) = (2u - 2u^2) \cdot \frac{(2u^2 - 2u + 1)}{(2u^2 - 2u)^{2}} = \frac{1}{2}\left(\frac{1}{u} + \frac{1}{1 - u} - 2\right).$$
Therefore, the maximum is attained at $1$.
I wasn't able to get anything with this, other than the fact that the numerator is sort of the derivative of the denominator. Also, the denominator is a geometric series. The substitution $u = (a + b + c)^{2} + (d + e + f)^{2}$ helped me rewrite things a bit, and it made my observations more clear.
Let $x=(a+b+c)^2+(d+e+f)^2$. By assumption, $0<x<1$. Therefore $$ f(a,b,c,d,e,f) =\frac{\sum_{n=0}^\infty nx^n}{\sum_{n=0}^\infty x^n} =\frac{x\frac d{dx}\sum_{n=0}^\infty x^n}{\sum_{n=0}^\infty x^n} =\frac{x/(1-x)^2}{1/(1-x)} =\frac{x}{1-x}, $$ which is not bounded above when $x\to1^-$. In particular, $\lim_{t\to0^+}f(1-t,0,0,t,0,0)=+\infty$.