Fractional inequality proof: If a/b < c/d, show a/b < (a+c)/(b+d) < c/d

john.s

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Hi folks, I’m going through a precalc book as a refresher and I guess my brain not operating as good as it used to. Got stuck on the following:

If a/b < c/d
Show a/b < (a+c)/(b+d) < c/d

Hint would be appreciated!

Thanks
 
Well, you cannot prove it because it is not invariably true. For example,

[MATH]\text {LET } a= 1, \ b = -\ 2, \ c = 3, \text { and } d = 10.[/MATH]
[MATH]\therefore \dfrac{a}{b} = -\ 0.5, \ \dfrac{a + c}{b + d} = \dfrac{1 + 3}{-\ 2 + 10} = 0.5, \text { and } \dfrac{c}{d} = 0.3.[/MATH]
[MATH]\therefore \dfrac{a + c}{b + d} \not < \dfrac{c}{d} \text { even though } \dfrac{a}{b} < \dfrac{c}{d}.[/MATH]
Or did you leave something out of the specification of the problem?
 
Thanks for the replies. I did forget to include that a,b,c,d are all positive numbers.
 
Hi folks, I’m going through a precalc book as a refresher and I guess my brain not operating as good as it used to. Got stuck on the following:

If a/b < c/d
Show a/b < (a+c)/(b+d) < c/d

Hint would be appreciated!

Thanks
If a/b < c/d

Multiply both sides by 'b/c' - what do you get?
 
If a/b < c/d
Show a/b < (a+c)/(b+d) < c/d Hint would be appreciated!
Thanks for the replies. I did forget to include that a,b,c,d are all positive numbers.
Rearrange & add one to both sides:
\(\displaystyle \dfrac{a}{c}<\dfrac{b}{d}\\\dfrac{a}{c}+1<\dfrac{b}{d}+1\\\dfrac{a+c}{c}<\dfrac{d+b}{d}\)
Continue
 
If a/b < c/d
Show a/b < (a+c)/(b+d) < c/d

Hint would be appreciated!

Use this fact \(\displaystyle \dfrac{a}{b} < \dfrac{c}{d} \ \implies \ ad < bc\)


Someone could split up the double inequality:

\(\displaystyle \dfrac{a}{b} \ \ vs. \ \ \dfrac{a + c}{b + d} \ \implies \ \)

\(\displaystyle a(b + d) \ \ vs. \ \ b(a + c) \ \implies\)

\(\displaystyle ab + ad \ \ vs. \ \ ab + bc \ \implies \ ?\)


and


\(\displaystyle \dfrac{a + c}{b + d} \ \ vs. \ \ \dfrac{c}{d} \ \implies \ \)

\(\displaystyle d(a + c) \ \ vs. \ \ c(b + d) \ \implies\)

\(\displaystyle ad + cd \ \ vs. \ \ \ bc + cd \ \implies \ ?\)
 
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Sorry for the late reply but many thanks for this hint. That does make for a very quick and simple proof.
 
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