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Help me please to prove the follow inequality: $\frac{a}{b}<\frac{a+k}{b+k},(a<b, a,b,k>0)$ thanky very much

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We have:$$a<b/\cdot k$$ $$ak<bk/+ab$$ $$ak+ab<bk+ab$$ $$a(b+k)<b(a+k)$$ $$\frac{a}{b}<\frac{a+k}{b+k}$$

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  • $\begingroup$ for me as I do not understand very well inequalities seems understandable with the above solution, thanks sir $\endgroup$ – user145717 May 1 '14 at 17:40
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HINT:

$$\frac{a+k}{b+k}-\frac ab=\frac{k(b-a)}{(b+k)b}$$ which is $>0$ as $b-a,k,b$ consequently $b+k$ are positive

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We divide the left term of the inequality for b:

$$\frac{a}{b}<\frac{\frac{a}{b}+\frac{k}{b}}{1+\frac{k}{b}}$$

putting $x:=a/b< 1$ and $y=k/b>0$:

$$x<\frac{x+y}{1+y}$$

whence

$$x+xy<x+y$$

But $xy<y$ because $x<1$ and $y>0$.

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