Ta có :
\(\left(a-b\right)\ge0\)
\(\Rightarrow a^2-2ab+b^2\ge0\)
\(\Rightarrow a^2+b^2\ge2ab\)
\(\Rightarrow a^2+b^2+2ab\ge4ab\)
\(\Rightarrow\left(a+b\right)^2\ge4ab\)
mà a,b>0
\(\Rightarrow\dfrac{\left(a+b\right)^2}{ab\left(a+b\right)}\ge\dfrac{4ab}{ab\left(a+b\right)}\)
\(\Rightarrow\dfrac{a+b}{ab}\ge\dfrac{4}{a+b}\)
\(\Rightarrow\dfrac{1}{a}+\dfrac{1}{b}\ge\dfrac{4}{a+b}\)
\(\Rightarrow\dfrac{1}{2}.\left(\dfrac{1}{a}+\dfrac{1}{b}\right)\ge\dfrac{1}{2}.\dfrac{4}{a+b}\)
\(\Rightarrow\dfrac{1}{2a}+\dfrac{1}{2b}\ge\dfrac{2}{a+b}\)
A/p bđt Cauchy-Schwarz dạng Engel có:
\(\dfrac{1}{2a}+\dfrac{1}{2b}\ge\dfrac{\left(1+1\right)^2}{2a+2b}=\dfrac{4}{2\left(a+b\right)}=\dfrac{2}{a+b}\left(đpcm\right)\)
