How to prove $\frac{a^2}{a-1} + \frac{b^2}{b+1} >= 1/3 $ when $a, b \in \mathbb{R}_{> 0}$ such that $a + b = 1$.

Question

Attempt: \begin{align*} \frac{a^2}{a-1}+\frac{b^2}{b+1} & = \frac{(a+b)(ab + a - b)}{(a-1)(b+1)}\\ & = \frac{ab + a - b}{(a-1)(b+1)} \\ & = \frac{ab+a-b}{(ab+a-b-1)} \\ & = \frac{a^2 - 3a +1}{a^2 - 3a + 2} \end{align*} Where do I go from here? A solution without the use of calculus would be preferred, but any help would be appreciated.

Answer 1

$b=1-a$, so the expression is $\dfrac {a^2}{a+1} + \dfrac{(1-a)^2}{2-a}$.

Over a common denominator, this simplifies to $\dfrac{a^2-a+1}{(2-a)(a+1)}=\dfrac{\left(a-\frac12\right)^2+\frac34}{-\left(a-\frac12\right)^2+\frac94}$.

Since $0\lt a\lt 1$, the numerator and denominator are both positive.

The numerator is minimized when $a=\frac12$, and the denominator is maximized when $a=\frac12$.

Therefore, the fraction is minimized when $a=\frac12$, and it is then $\dfrac{\frac34}{\frac94}=\dfrac13$.