Từ BĐT trên ,ta có:
\(\dfrac{1}{a}\)+\(\dfrac{1}{b}\) \(\geq\) \(\dfrac{4}{a+b}\)
\(\Leftrightarrow\) \(\dfrac{a+b}{ab}\) \(\geq\) \(\dfrac{4}{a+b}\)
\(\Leftrightarrow\) (a+b)(a+b) \(\geq\) 4ab
\(\Leftrightarrow\) (a+b)2 \(\geq\) 4ab
\(\Leftrightarrow\) a2 +2ab+b2\(\geq\) 4ab
\(\Leftrightarrow\) a2+2ab+b2-4ab \(\geq\) 0
\(\Leftrightarrow\) a2-2ab+b2 \(\geq\) 0
\(\Leftrightarrow\) (a-b)2 \(\geq\) 0 (luôn đúng)
Dấu '=' xảy ra khi và chỉ khi a=b
Từ đó ta chứng minh được BĐT : \(\dfrac{1}{a}\) +\(\dfrac{1}{b}\)\(\geq\) \(\dfrac{4}{a+b}\)
\(\dfrac{1}{a}+\dfrac{1}{b}=\dfrac{a+b}{ab}=\dfrac{\left(a+b\right)^2}{ab\left(a+b\right)}\) (1)
\(\dfrac{4}{a+b}=\dfrac{4ab}{ab\left(a+b\right)}\) (2)
ta có:
\(\left(a+b\right)^2\ge\left(a-b\right)^2\) và \(\left(a-b\right)^2\ge4ab\)
nên \(\left(a+b\right)^2\ge4ab\) (3)
từ (1), (2) và (3) suy ra \(\dfrac{\left(a+b\right)^2}{ab\left(a+b\right)}\ge\dfrac{4ab}{ab\left(a+b\right)}\) hay \(\dfrac{1}{a}+\dfrac{1}{b}\ge\dfrac{4}{a+b}\)(đpcm)
Xét hiệu : VT - VP
= \(\dfrac{1}{a}\) + \(\dfrac{1}{b}\) - \(\dfrac{4}{a+b}\) = \(\dfrac{a+b}{ab}\) - \(\dfrac{4}{a+b}\)
= \(\dfrac{\left(a+b\right)^2}{ab.\left(a+b\right)}\) - \(\dfrac{4ab}{ab.\left(a+b\right)}\)
= \(\dfrac{a^2+2ab+b^2-4ab}{ab.\left(a+b\right)}\) = \(\dfrac{\left(a-b\right)^2}{ab.\left(a+b\right)}\)
Vì a,b > 0 => a.b > 0 ; a + b > 0
=> ab.(a + b) > 0 . Mà ( a - b)2 > 0 nên
\(\dfrac{\left(a-b\right)^2}{ab.\left(a+b\right)}\) \(\ge\) 0 hay VT - VP \(\ge\) 0 ( BĐT đúng )
=> \(\dfrac{1}{a}\) + \(\dfrac{1}{b}\) \(\ge\) \(\dfrac{4}{a+b}\)
