Giả sử \(a\ge b\) suy ra a = b + m (m \(\ge\) 0).
Ta có \(\frac{a}{b}+\frac{b}{a}=\frac{b+m}{b}+\frac{b}{b+m}\)
\(=\frac{b}{b}+\frac{m}{b}+\frac{b}{b+m}=1+\frac{m}{b}+\frac{b}{b+m}\ge1+\frac{m}{b+m}+\frac{b}{b+m}=1+\frac{m+b}{b+m}\)
\(=1+1=2\)
Vậy \(\frac{a}{b}+\frac{b}{a}\ge2\) (dấu = xảy ra \(\Leftrightarrow\) m = 0 \(\Leftrightarrow\) a = b)
1 đ-ú-n-g nha, nghĩ mãi mới ra đó !
Ta có:
\(\frac{a}{b}>0\Rightarrow a,b\ne0\)
Giả sử: \(a\ge b\)Đặt: \(a=b+m\left(m\in N\right)\Rightarrow\frac{b+m}{b}+\frac{b}{b+m}=\frac{a}{b}+\frac{b}{a}\)
\(=1+\frac{m}{b}+1-\frac{m}{b+m}=2+\frac{m}{b}-\frac{m}{b+m}\) Vì: \(b\le b+m\Rightarrow\frac{m}{b}\ge\frac{m}{b+m}\Rightarrow\frac{a}{b}+\frac{b}{a}\ge2\left(ĐPCM\right)\)
Cách lớp 7 nha!
Đặt \(\sqrt{\frac{a}{b}}=t\left(t>0\right)\).Ta cần c/m \(t^2+\frac{1}{t^2}\ge2\)
Thật vậy,\(\left(t-\frac{1}{t}\right)^2\ge0\Leftrightarrow t^2-2+\frac{1}{t^2}\ge0\)
\(\Leftrightarrow t^2+\frac{1}{t^2}\ge2^{\left(đpcm\right)}\)
Dấu "=" xảy ra khi \(t=\frac{1}{t}\Leftrightarrow t=1\Leftrightarrow a=b\)
