\(\dfrac{1}{a}+\dfrac{1}{b}\ge\dfrac{4}{a+b}\)
\(\Leftrightarrow\dfrac{a+b}{ab}\ge\dfrac{4}{a+b}\)
\(\Leftrightarrow\left(a+b\right)^2\ge4ab\)
\(\Leftrightarrow a^2+2ab+b^2-4ab\ge0\)
\(\Leftrightarrow\left(a-b\right)^2\ge0\)
BĐT đúng .
Áp dụng BĐT Cô - si dạng Engel , ta có :
\(\dfrac{1}{a}+\dfrac{1}{b}\) ≥ \(\dfrac{\left(1+1\right)^2}{a+b}=\dfrac{4}{a+b}\)
Đẳng thức xảy ra khi : a = b
1/a+1/b>_4/a+b
<=>a+b/ab>_4/a+b
<=>(a+b)^2>_4a<=>a^2+2ab+b^2-4ab>_0
<=>(a-b)^2>_0=>BDT dung