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easy inequality

I am trying to show that if 0<a<1 and 0<b<1 and we know that a is less than b then:

2ab+ab+a+2ab<12 \frac{2ab+a}{b+a+2ab} < \frac{1}{2}

I have simplified the fractions to:

2b+11+2b+ba\frac{2b+1}{1+2b+\frac{b}{a}}

but I can't see how this is less than 0.5
Reply 1
Avatar for nuodai
nuodai
17
Your best bet is to multiply through by the denominators of the two fractions to get 2(2ab+a)<b+a+4ab2(2ab+a) < b+a+4ab (we can do this since everything is positive). Why must this be true?
(edited 13 years ago)
Reply 2
Avatar for sonic7899
sonic7899
OP
That must be true because b<1?

but is that not using the fact that it is less than a half without having shown it?
2ab + a < b

Should be false. Take a = 1/2 and b arbitrary in (1/2,1).
Reply 4
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sonic7899
OP
so do you mean that it the inequality in the op is not true?
Original post by sonic7899
so do you mean that it the inequality in the op is not true?


Unless 1/2 < 0 it's false.
Reply 6
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sonic7899
OP
Original post by DeanK22
Unless 1/2 < 0 it's false.


Erm I think I made a mistake in my workings. The fraction I should be showing is less than a half is actually:


2ab+a4ab+a+b\frac{2ab+a}{4ab+a+b}
Reply 7
Avatar for nuodai
nuodai
17
Original post by sonic7899
Erm I think I made a mistake in my workings. The fraction I should be showing is less than a half is actually:


2ab+a4ab+a+b\frac{2ab+a}{4ab+a+b}


That makes more sense. You need to use the fact that b>ab>a.
Original post by sonic7899
Erm I think I made a mistake in my workings. The fraction I should be showing is less than a half is actually:


2ab+a4ab+a+b\frac{2ab+a}{4ab+a+b}


In which case, nuodai's suggestion is the way to go here. What is the original question?
Reply 9
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sonic7899
OP
Original post by nuodai
That makes more sense. You need to use the fact that b>ab>a.


Ah I think I have just got it. Is is just that a<b and so:

2ab+a4ab+a+b<2ab+a4ab+a+a=2ab+a2[2ab+a]=12\frac{2ab+a}{4ab+a+b} < \frac{2ab+a}{4ab+a+a} = \frac{2ab+a}{2[2ab+a]} = \frac{1}{2}?
Reply 10
Avatar for nuodai
nuodai
17
Original post by sonic7899
Ah I think I have just got it. Is is just that a<b and so:

2ab+a4ab+a+b<2ab+a4ab+a+a=2ab+a2[2ab+a]=12\frac{2ab+a}{4ab+a+b} < \frac{2ab+a}{4ab+a+a} = \frac{2ab+a}{2[2ab+a]} = \frac{1}{2}?


Seems fine to me :smile:

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