\(Từ GT, ta có: \(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}+\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ca}\ge6\) Áp dụng bđt AM - GM, ta lại có: \(\frac{1}{a^2}+1\ge\frac{2}{a};\frac{1}{b^2}+1\ge\frac{2}{b};\frac{1}{c^2}+1\ge\frac{2}{c}\) \(\frac{1}{a^2}+\frac{1}{b^2}\ge\frac{2}{ab};\frac{1}{b^2}+\frac{1}{c^2}\ge\frac{2}{bc};\frac{1}{c^2}+\frac{1}{a^2}\ge\frac{2}{ac}\) Cộng theo vế ta có: \(3\left(\text{∑}\frac{1}{a^2}\right)+3\ge2\left(\text{∑}\frac{1}{a}+\text{∑}\frac{1}{ab}\right)\Leftrightarrow\text{∑}\frac{1} {a^2}\ge3\left(đ\text{pcm}\right)\) \(\text{Dau }"="\Leftrightarrow a=b=c=1\)
Từ GT, ta có: \(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}+\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ca}\ge6\)
Áp dụng bđt AM - GM, ta lại có:
\(\frac{1}{a^2}+1\ge\frac{2}{a};\frac{1}{b^2}+1\ge\frac{2}{b};\frac{1}{c^2}+1\ge\frac{2}{c}\)
\(\frac{1}{a^2}+\frac{1}{b^2}\ge\frac{2}{ab};\frac{1}{b^2}+\frac{1}{c^2}\ge\frac{2}{bc};\frac{1}{c^2}+\frac{1}{a^2}\ge\frac{2}{ac}\)
Cộng theo vế ta có:
\(3\left(\text{∑}\frac{1}{a^2}\right)+3\ge2\left(\text{∑}\frac{1}{a}+\text{∑}\frac{1}{ab}\right)\Leftrightarrow\text{∑}\frac{1}{a^2}\ge3\left(đ\text{pcm}\right)\)
\(\text{Dau }"="\Leftrightarrow a=b=c=1\)
