Question

Cho a, b là số dương . chứng minh rằng :\(\frac{1}{a}+\frac{1}{b}\ge\frac{4}{a+b}\)

GIẢI GIÚP MÌNH VỚI MAI MÌNH ĐI HỌC RỒI

 

Answer 1

\(Để\frac{1}{a}+\frac{1}{b}\ge\frac{4}{a+b}\)

\(\Leftrightarrow\frac{1}{a}+\frac{1}{b}-\frac{4}{a+b}\ge0\)

\(\Leftrightarrow\frac{a+b}{ab}-\frac{4}{a+b}\ge0\)

\(\Leftrightarrow\frac{\left(a+b\right)^2}{ab\left(a+b\right)}-\frac{4ab}{ab\left(a+b\right)}\ge0\)

\(\Leftrightarrow\frac{a^2+2ab+b^2-4ab}{ab\left(a+b\right)}\ge0\)

\(\Leftrightarrow\frac{a^2-2ab+b^2}{ab\left(a+b\right)}\ge0\)

\(\Leftrightarrow\frac{\left(a-b\right)^2}{ab\left(a+b\right)}\ge0\left(đpcm\right)\)

Vậy \(\frac{1}{a}+\frac{1}{b}\ge\frac{4}{a+b}\)

Answer 2

\(\frac{1}{a}+\frac{1}{b}\ge\frac{4}{a+b}\)

\(\Leftrightarrow\frac{a+b}{ab}\ge\frac{4}{a+b}\)

\(\Leftrightarrow\left(a+b\right)^2\ge4ab\)

\(\Leftrightarrow\left(a+b\right)^2-4ab\ge0\)

\(\Leftrightarrow\left(a-b\right)^2\ge0\)(Luôn đúng)