Áp dụng BĐT Cô - si : x + y ≥ \(2\sqrt{xy}\) ( x > 0 ; y > 0)
⇒ \(\dfrac{1}{a}+\dfrac{1}{b}\) ≥ \(\dfrac{2}{\sqrt{ab}}\) ( a > 0 ; b > 0 )
⇒ ( a + b)\(\left(\dfrac{1}{a}+\dfrac{1}{b}\right)\) ≥ \(\dfrac{2}{\sqrt{ab}}\).\(2\sqrt{ab}\)
⇒ ( a + b)\(\left(\dfrac{1}{a}+\dfrac{1}{b}\right)\) ≥ 4
Xét hiệu:
\(\left(a+b\right)\left(\dfrac{1}{a}+\dfrac{1}{b}\right)-4=1+\dfrac{a}{b}+\dfrac{b}{a}+1-4\)
\(=\dfrac{a}{b}+\dfrac{b}{a}-2=\dfrac{a^2+b^2-2ab}{ab}=\dfrac{\left(a-b\right)^2}{ab}\ge0\) ( luôn đúng)
Suy ra: \(\left(a+b\right)\left(\dfrac{1}{a}+\dfrac{1}{b}\right)\ge4\)
*đây là cách không áp dụng bất đẳng thức nào nhé*
Ta có: \(\left(a+b\right)\left(\dfrac{1}{a}+\dfrac{1}{b}\right)\)
\(=\dfrac{a}{a}+\dfrac{a}{b}+\dfrac{b}{a}+\dfrac{b}{b}\)
\(=2+\dfrac{a}{b}+\dfrac{b}{a}\)
Xét \(\dfrac{a}{b}+\dfrac{b}{a}-2=\dfrac{a^2-2ab+b^2}{ab}=\dfrac{\left(a-b\right)^2}{ab}\ge0\)
do \(\left\{{}\begin{matrix}\left(a-b\right)^2\ge0\forall a,b\\a>0,b>0\Rightarrow ab>0\end{matrix}\right.\)
\(\Rightarrow\dfrac{a}{b}+\dfrac{b}{a}-2\ge0\)
\(\Rightarrow\dfrac{a}{b}+\dfrac{b}{a}\ge2\)
Cộng 2 vế của \(\dfrac{a}{b}+\dfrac{b}{a}\ge2\) với 2, ta được
\(2+\dfrac{a}{b}+\dfrac{b}{a}\ge4\)
\(\Rightarrow\left(a+b\right)\left(\dfrac{1}{a}+\dfrac{1}{b}\right)\ge4\)
Theo BĐT Cô - si ta có :
\(\left\{{}\begin{matrix}a+b\ge2\sqrt{ab}\\\dfrac{1}{a}+\dfrac{1}{b}\ge2\sqrt{\dfrac{1}{ab}}\end{matrix}\right.\)
Nhân vế theo vế ta được :
\(\left(a+b\right)\left(\dfrac{1}{a}+\dfrac{1}{b}\right)\ge2\sqrt{ab}.2\sqrt{\dfrac{1}{ab}}=4\)
Dấu \("="\) xảy ra khi \(a=b\)
\(\left(a+b\right)\left(\dfrac{1}{a}+\dfrac{1}{b}\right)\ge4\)
Áp dụng BDT \(\dfrac{1}{x}+\dfrac{1}{y}\ge\dfrac{4}{x+y}\)
\(\Rightarrow\dfrac{1}{a}+\dfrac{1}{b}\ge\dfrac{4}{a+b}\\ \Rightarrow\left(a+b\right)\left(\dfrac{1}{a}+\dfrac{1}{b}\right)\ge\left(a+b\right)\cdot\dfrac{4}{a+b}\ge4\left(đpcm\right)\)
Vậy \(\left(a+b\right)\left(\dfrac{1}{a}+\dfrac{1}{b}\right)\ge4\)
đẳng thức xảy ra khi \(a=b\)