Help with fraction inequality Let $a,b,c$ be three numbers such that:

*

*$a,b\in (0.5,1)$

*$c \in (0.25,0.5)$

*$c < 0.5a$

*$c > 0.5b$

*$a + b < 1 + c$
Let
$$f(a,b,c) = \frac{1+c}{a+\frac{bc}{c+0.5b}}$$


*

*What is the best upper bound we can give of $f$?


*Can we show that $f\leq 1.5$? (can you find $a,b,c$ such that $f(a,b,c)>1.5$?)

Currently, my best upper bound is $\frac{15}{8}$, but it seems it's not close to being tight (see my proof below) as on one place I use the upper range (in the numerator) of $c$ and on different stage I use the lower range of $c$ (in the denominator).

My first attempt showed $f<2.5$:
$$f(a,b,c) = \frac{1+c}{a+\frac{bc}{c+0.5b}}<\frac{1+0.5a}{a}=0.5+\frac{1}{a}=2.5$$

Found Improvement to $f<\frac{8}{3}$:
$$f(a,b,c) = \frac{1+c}{a+\frac{bc}{c+0.5b}}<\frac{1+c}{a+\frac{0.5b^2}{0.5b+0.5b}}=\frac{1+c}{a+0.5b}<\frac{1+0.5a}{a+0.5b}<\frac{1+1}{0.5+0.5^2}=\frac{8}{3}$$

Further improvement to $f<\frac{15}{8}$ (using $c<b$ which is obvious by their domains):
$$f(a,b,c) = \frac{1+c}{a+\frac{bc}{c+0.5b}}<\frac{1+c}{2c+\frac{c^2}{c+0.5c}}$$
$$=\frac{1+c}{2c+\frac{2}{3}c}=\frac{1+c}{\frac{8}{3}c}=\frac{3}{8}+\frac{1}{\frac{8}{3}c}$$
$$=\frac{3}{8}+\frac{3}{8c}<\frac{3}{8}+\frac{12}{8}=\frac{15}{8}$$

We have $\frac12 < b < 2c < a < 1$. Hence:
$$\frac{1+c}{a+\frac{bc}{c+b/2}} < \frac{1+c}{a+\frac{bc}{2c}} < \frac{1+\frac12a}{\frac32a} = \frac2{3a}+\frac13 < \frac43+\frac13 = \frac53$$ 
Further as $b, 2c, a \to \frac12$, the the function comes arbitrarily close to $\frac53$, so this is indeed the best/lowest upper bound.
You already have cases where $f> 1.5$, e.g. $a=0.52, b = 0.51, c = \frac12(a+b)$... 
