Áp dụng bđt Bunhiacopxki ta có :
\(\left(1+1+1\right)\left[\left(a+\frac{1}{a}\right)^2+\left(b+\frac{1}{b}\right)^2+\left(c+\frac{1}{c}\right)^2\right]\ge\left(a+\frac{1}{a}+b+\frac{1}{b}+c+\frac{1}{c}\right)^2\)
\(\Leftrightarrow\left(a+\frac{1}{a}\right)^2+\left(b+\frac{1}{b}\right)^2+\left(c+\frac{1}{c}\right)^2\ge\frac{\left(a+b+c+\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)^2}{3}\)
\(=\frac{\left(1+\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)^2}{3}\)
Ta lại có : \(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\ge\frac{9}{a+b+c}\)(bđt quen thuộc; tự cm)
Nên \(\frac{\left(1+\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)^2}{3}\ge\frac{\left(1+\frac{9}{a+b+c}\right)^2}{3}=\frac{10^2}{3}=\frac{100}{3}>\frac{99}{3}=33\)
Hay \(\left(a+\frac{1}{a}\right)^2+\left(b+\frac{1}{b}\right)^2+\left(c+\frac{1}{c}\right)^2>33\)(đpcm)
