Question

Cho 3 số dương a,b,c và a+b+c\(\le\frac{3}{2}\). Chưng minh rằng

\(a+b+c+\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\ge\frac{15}{2}\)

Answer 1

Ta có:

\(a+b+c+\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\ge a+b+c+\frac{9}{a+b+c}\)

Đặt: \(a+b+c=t\le\frac{3}{2}\Leftrightarrow2t\le3\)

Ta có: Cần cm: \(t+\frac{9}{t}\ge\frac{15}{2}\Leftrightarrow\frac{t^2+9}{t}\ge\frac{15}{2}\Leftrightarrow2t^2+18-15t\ge0\)

\(\Leftrightarrow\left(2t^2-3t\right)+\left(18-12t\right)\ge0\Leftrightarrow t\left(2t-3\right)-6\left(2t-3\right)\ge0\Leftrightarrow\left(t-6\right)\left(2t-3\right)\ge0\)(đúng với \(t\le\frac{3}{2}\))

Dấu "=" khi \(a=b=c=\frac{1}{2}\)