Ta có :
\(\left(a-1\right)^2\ge0\)
\(\Rightarrow a^2+1\ge2a\)
\(\Rightarrow2a-1\le a^2\)
\(\Rightarrow\dfrac{a}{2a-1}\ge\dfrac{a}{a^2}\)
\(\Rightarrow\dfrac{a}{2a-1}\ge\dfrac{1}{a}\)
Tương tự ta có :
\(\dfrac{b}{2b-1}\ge\dfrac{1}{b}\)
Do đó : \(\dfrac{a}{2a-1}+\dfrac{b}{2b-1}\ge\dfrac{1}{a}+\dfrac{1}{b}\)
*) Chứng minh bổ đề : \(\dfrac{1}{x}+\dfrac{1}{y}\ge\dfrac{4}{x+y}\left(x,y>0\right)\)
Ta có :
\(\left(x-y\right)^2\ge0\)
\(\Rightarrow x^2+y^2\ge2xy\)
\(\Rightarrow x^2+y^2+2xy\ge4xy\)\(\Rightarrow\left(x+y\right)^2\ge4xy\)
\(\Rightarrow\dfrac{\left(x+y\right)^2}{xy\left(x+y\right)}\ge\dfrac{4xy}{xy\left(x+y\right)}\left(x,y>0\right)\)
\(\Rightarrow\dfrac{x+y}{xy}\ge\dfrac{4}{x+y}\)
\(\Rightarrow\dfrac{1}{x}+\dfrac{1}{y}\ge\dfrac{4}{x+y}\)
*) Áp dụng bổ đè trên ta có:
\(\dfrac{a}{2a-1}+\dfrac{b}{2b-1}\ge\dfrac{1}{a}+\dfrac{1}{b}\ge\dfrac{4}{a+b}\) (1)
Ta có :
\(a+b-\left(1+ab\right)\)
\(=\left(a-ab\right)+\left(b-1\right)\)
\(=a\left(1-b\right)+\left(b-1\right)\)
\(=\left(1-b\right)\left(a-1\right)\)
Vì \(a,b\ge1\Rightarrow\left\{{}\begin{matrix}a-1\ge0\\1-b\le0\end{matrix}\right.\)
\(\Rightarrow\left(1-b\right)\left(a-1\right)\le0\)
\(\Rightarrow a+b-\left(1+ab\right)\le0\)
\(\Rightarrow a+b\le1+ab\)
\(\Rightarrow\dfrac{4}{a+b}\ge\dfrac{4}{1+ab}\) (2)
Từ (1) và (2) ta được:
\(\dfrac{a}{2a-1}+\dfrac{b}{2b-1}\ge\dfrac{1}{a}+\dfrac{1}{b}\ge\dfrac{4}{a+b}\ge\dfrac{4}{1+ab}\)
\(\Rightarrow\dfrac{a}{2a-1}+\dfrac{b}{2b-1}\ge\dfrac{4}{1+ab}\)
\(\rightarrowđpcm\)
