If $b \geq a$ . $\implies \frac{b}{a} \geq 1$ $\implies \frac{a+1}{b+c} \geq 1$ $\implies a + 1 \geq b + c \geq a + c$ $\implies c \leq 1$ Similarly $c \geq 1$ if $b \leq a$. So $c \geq 1$ doesn't hold always.

The given is equivalent to $c = \frac{a^2 + a - b^2}{b}$. If $a \ge b$, then this is obvious. So suppose $a < b \implies b \ge a+1$, but then $c \le \frac{a^2 + a - (a+1)^2)}{b}$, which is negative, which is impossible since $c$ was a positive real number

Suppose $c<1$. Note that this implies $$\frac{b}{a}=\frac{a+1}{b+c} > \frac{a+1}{b+1}.$$ which is equivalent to $b^2+b>a^2+a \iff (b-a)(b+a+1)>0$. So this can only be true if $b>a$. Additionally $$\frac{a+1}{b} > \frac{a+1}{b+c}=\frac{b}{a}$$ must hold because $c>0$. But because $b>a$ it follows that $\frac{b}{a}>1$ which leads to $\frac{a+1}{b}>1$ with the above. So $a+1>b>a$ must be true but this is a contradiction to $a, b \in \mathbb{Z}^+$. Therefore $c \geq 1$.