Ta sẽ CM : \(\dfrac{a}{b}>\dfrac{a+m}{b+m}\left(a;b;m>0;a>b\right)\)
Thật vậy ; ta có :
\(a>b\\ \Rightarrow am>bm\\ \Rightarrow ab+am>ab+bm\\ \Rightarrow a\left(b+m\right)>b\left(a+m\right)\\ \Rightarrow\dfrac{a}{b}>\dfrac{a+m}{b+m}\left(đpcm\right)\)
Áp dụng BĐT trên ; có :
\(\dfrac{2^{2012}+1}{2^{2009}+1}>\dfrac{2^{2012}+1+3}{2^{2009}+1+3}\\ =\dfrac{2^{2012}+2^2}{2^{2009}+2^2}\\ =\dfrac{2^2\left(2^{2010}+1\right)}{2^2\left(2^{2007}+1\right)}\\ =\dfrac{2^{2010}+1}{2^{2007}+1}\)