Ta có : \(\left(a-b\right)^2\ge0\)
\(\Leftrightarrow a^2-2ab+b^2\ge0\)
\(\Leftrightarrow a^2+2ab+b^2\ge4ab\)
\(\Leftrightarrow\left(a+b\right)^2\ge4ab\)
\(\Leftrightarrow\frac{a+b}{ab}\ge\frac{4}{a+b}\)
\(\Leftrightarrow\frac{1}{a}+\frac{1}{b}\ge\frac{4}{a+b}\) (1)
Ta cũng có :
\(-\left(a-b\right)^2\le0\)
\(\Leftrightarrow-a^2+2ab-b^2\le0\)
\(\Leftrightarrow a^2+2ab+b^2\le2\left(a^2+b^2\right)\)
\(\Leftrightarrow\left(a+b\right)^2\le2\left(a^2+b^2\right)\)
\(\Leftrightarrow\frac{16}{\left(a+b\right)^2}\ge\frac{16}{2\left(a^2+b^2\right)}\)
\(\Leftrightarrow\frac{16}{\left(a+b\right)^2}\ge\frac{8}{a^2+b^2}\)
\(\Leftrightarrow\sqrt{\frac{16}{\left(a+b\right)^2}}\ge\sqrt{\frac{8}{a^2+b^2}}\)
\(\Rightarrow\frac{4}{a+b}\ge\frac{2\sqrt{2}}{\sqrt{a^2+b^2}}\) (2)
Từ (1) ; (2) \(\Rightarrow\frac{1}{a}+\frac{1}{b}\ge\frac{4}{a+b}\ge\frac{2\sqrt{2}}{\sqrt{a^2+b^2}}\) (đpcm)