# If $\frac{a}{b}<\frac{c}{d}$ and $\frac{e}{f}<\frac{g}{h}$, then $\frac{a+e}{b+f} < \frac{c+g}{d+h}$.

If $$a, b, c, d, e, f, g, h$$ are positive numbers satisfying $$\frac{a}{b}<\frac{c}{d}$$ and $$\frac{e}{f}<\frac{g}{h}$$ and $$b+f>d+h$$, then $$\frac{a+e}{b+f} < \frac{c+g}{d+h}$$.

I thought it is easy to prove. But I could not. How to prove this? Thank you.

The question is a part of a bigger proof I am working on. There are two strictly concave, positive valued, strictly increasing functions $$f_1$$ and $$f_2$$ (See Figure 1). Given 4 points $$x_1$$, $$x_2$$, $$x_3$$ and $$x_4$$ such that $$x_1< x_i$$, $$i=2, 3,4$$ and $$x_4> x_i$$, $$i=1, 2, 3$$, let $$d=x_2-x_1$$, $$b=x_4-x_3$$ $$c=f_1(x_2)-f_1(x_1)$$, $$a=f_1(x_4)-f_1(x_3)$$. And given 4 points $$y_1$$, $$y_2$$, $$y_3$$ and $$y_4$$ such that $$y_1< y_i$$, $$i=2, 3,4$$ and $$y_4> y_i$$, $$i=1, 2, 3$$, let $$h=y_2-y_1$$, $$f=y_4-y_3$$ $$g=f_2(y_2)-f_2(y_1)$$, $$e=f_2(y_4)-f_2(y_3)$$.

Since the functions are concave, we have $$\frac{a}{b}<\frac{c}{d}$$ and $$\frac{e}{f}<\frac{g}{h}$$. And I thought in this setting, it is true that $$\frac{a+e}{b+f} < \frac{c+g}{d+h}$$ even without the restriction $$b+f>d+h$$. • I added a new restriction $b+f>d+h$. Refer to Keith's nice counter example for the original posting.
– Sang
Jul 9, 2015 at 12:24
• Your unlucky numbers are $a,b,c,d,e,f,g,h = 1, 1, 6, 5, 10, 7, 2, 1$. May I ask you what the original purpose of this question is? (I suppose it is something related to Farey series.) Jul 9, 2015 at 12:26
• Hi Darij, it is part of a proof I am struggling with. I tried to describe what I am doing by adding more to the original question. Please refer to the edited question. Thank you.
– Sang
Jul 9, 2015 at 13:32
• This is a great example of something that seems obvious but is actually demonstrably false. In statistics this phenomenon is well-known by the name of Simpson's paradox. Jul 9, 2015 at 14:34

This is false.

For example, $$\frac{1}{3} < \frac{5}{12}, \quad \frac{52}{5} < \frac{11}{1}, \quad \frac{53}{8} > \frac{16}{13}.$$

• Is there a counterexample that uses fractions in lowest terms? Jul 8, 2015 at 16:06
• @GTonyJacobs I've edited my answer. Jul 8, 2015 at 16:09
• Nice counterexample. Jul 8, 2015 at 16:15
• It is a very nice counterexample. Is there a counter example that satisfies an added restriction, $b+f>d+h$? Or the statement is true with the added restriction?
– Sang
Jul 9, 2015 at 12:20
• @Sang The extra restriction is irrelevant: just rescale $a,b,e,f$ by the same large integer and you'll get $b+f > d+h$ without changing any of the values being compared. You could then add $1$ to $b$ and $f$ and have a decent chance of the resulting fractions being in lowest-terms (certainly high enough that you could just pick a larger integer that works): these are strict inequalities so they are tolerant of minute adjustments. Jul 9, 2015 at 14:31

The updated question (with the additional constraint $b+f>d+h$) is also false. For example,

$\frac{1}{1}<\frac{3}{2}$ and $\frac{9}{4}<\frac{5}{2}$, but $\frac{1+9}{1+4} = \frac{10}{5} = \frac{8}{4} = \frac{3+5}{2+2}$.

As

$$\frac{a+e}{b+f} < \frac{c+g}{d+h}$$ $$(a+e)(d+h) < (c+g)(b+f)$$ $$ad+eh+ah+ed < cb+fg+cf+gb$$ here put $ah+ed = x$ and $cf+gb = y$. Now we have $$ad+eh+x < bc+fg+y$$ Now as the given $$\frac{a}{b}<\frac{c}{d} \implies { ad<bc}$$ and $$\frac{e}{f}< \frac{g}{h} \implies{eh<fg}$$ so by adding both the above eqs $$ad+eh<bc+fg$$ and by this condition we have $$ad+eh+x<bc+fg+y$$

• How do you justify the last step? It is certainly true if $x<y$ (in particular for $\frac{a}{b}<\frac{g}{h}$ and $\frac{c}{d}<\frac{e}{f}$), but we cannot assume that. Jul 9, 2015 at 12:50