By using AM-GM ( x 2 + y 2 + z 2 ≥ x y + y z + z x ) \left(x^{2}+y^{2}+z^{2} \geq x y+y z+z x\right) ( x 2 + y 2 + z 2 ≥ x y + y z + z x )
( a + 1 b ) 2 + ( b + 1 c ) 2 + ( c + 1 a ) 2 ≥ ( a + 1 b ) ( b + 1 c ) + ( b + 1 c ) ( c + 1 a ) + ( c + 1 a ) ( a + 1 b ) = ( a b + 1 + a c + a ) + ( b c + 1 + b a + b ) + ( c a + 1 + c b + c ) = a b + b c + c a + a c + c b + b a + 3 + a + b + c
\begin{aligned}
\left(a+\frac{1}{b}\right)^{2}+\left(b+\frac{1}{c}\right)^{2}+\left(c+\frac{1}{a}\right)^{2} & \geq\left(a+\frac{1}{b}\right)\left(b+\frac{1}{c}\right)+\left(b+\frac{1}{c}\right)\left(c+\frac{1}{a}\right)+\left(c+\frac{1}{a}\right)\left(a+\frac{1}{b}\right) \\
& =\left(a b+1+\frac{a}{c}+a\right)+\left(b c+1+\frac{b}{a}+b\right)+\left(c a+1+\frac{c}{b}+c\right) \\
& =a b+b c+c a+\frac{a}{c}+\frac{c}{b}+\frac{b}{a}+3+a+b+c
\end{aligned}
( a + b 1 ) 2 + ( b + c 1 ) 2 + ( c + a 1 ) 2 ≥ ( a + b 1 ) ( b + c 1 ) + ( b + c 1 ) ( c + a 1 ) + ( c + a 1 ) ( a + b 1 ) = ( ab + 1 + c a + a ) + ( b c + 1 + a b + b ) + ( c a + 1 + b c + c ) = ab + b c + c a + c a + b c + a b + 3 + a + b + c
Notice that by AM-GM we have a b + b a ≥ 2 b , b c + c b ≥ 2 c a b+\frac{b}{a} \geq 2 b, b c+\frac{c}{b} \geq 2 c ab + a b ≥ 2 b , b c + b c ≥ 2 c c a + a c ≥ 2 a c a+\frac{a}{c} \geq 2 a c a + c a ≥ 2 a
Thus,
( a + 1 b ) 2 + ( b + 1 c ) 2 + ( c + 1 a ) 2 ≥ ( a b + b a ) + ( b c + c b ) + ( c a + a c ) + 3 + a + b + c ≥ 3 ( a + b + c + 1 )
\left(a+\frac{1}{b}\right)^{2}+\left(b+\frac{1}{c}\right)^{2}+\left(c+\frac{1}{a}\right)^{2} \geq\left(a b+\frac{b}{a}\right)+\left(b c+\frac{c}{b}\right)+\left(c a+\frac{a}{c}\right)+3+a+b+c \geq 3(a+b+c+1)
( a + b 1 ) 2 + ( b + c 1 ) 2 + ( c + a 1 ) 2 ≥ ( ab + a b ) + ( b c + b c ) + ( c a + c a ) + 3 + a + b + c ≥ 3 ( a + b + c + 1 )
The equality holds if and only if a = b = c = 1 a=b=c=1 a = b = c = 1
Solution2. From QM-AM we obtain
( a + 1 b ) 2 + ( b + 1 c ) 2 + ( c + 1 a ) 2 3 ≥ a + 1 b + b + 1 c + c + 1 a 3 ⇔ ( a + 1 b ) 2 + ( b + 1 c ) 2 + ( c + 1 a ) 2 ≥ ( a + 1 b + b + 1 c + c + 1 a ) 2 3
\begin{aligned}
& \sqrt{\frac{\left(a+\frac{1}{b}\right)^{2}+\left(b+\frac{1}{c}\right)^{2}+\left(c+\frac{1}{a}\right)^{2}}{3}} \geq \frac{a+\frac{1}{b}+b+\frac{1}{c}+c+\frac{1}{a}}{3} \Leftrightarrow \\
& \left(a+\frac{1}{b}\right)^{2}+\left(b+\frac{1}{c}\right)^{2}+\left(c+\frac{1}{a}\right)^{2} \geq \frac{\left(a+\frac{1}{b}+b+\frac{1}{c}+c+\frac{1}{a}\right)^{2}}{3}
\end{aligned}
3 ( a + b 1 ) 2 + ( b + c 1 ) 2 + ( c + a 1 ) 2 ≥ 3 a + b 1 + b + c 1 + c + a 1 ⇔ ( a + b 1 ) 2 + ( b + c 1 ) 2 + ( c + a 1 ) 2 ≥ 3 ( a + b 1 + b + c 1 + c + a 1 ) 2
From AM-GM we have 1 a + 1 b + 1 c ≥ 3 1 a b c 3 = 3 \frac{1}{a}+\frac{1}{b}+\frac{1}{c} \geq 3 \sqrt[3]{\frac{1}{a b c}}=3 a 1 + b 1 + c 1 ≥ 3 3 ab c 1 = 3
( a + 1 b ) 2 + ( b + 1 c ) 2 + ( c + 1 a ) 2 ≥ ( a + 1 b + b + 1 c + c + 1 a ) 2 3 ≥ ( a + b + c + 3 ) 2 3 = = ( a + b + c ) ( a + b + c ) + 6 ( a + b + c ) + 9 3 ≥ ( a + b + c ) 3 a b c 3 + 6 ( a + b + c ) + 9 3 = = 9 ( a + b + c ) + 9 3 = 3 ( a + b + c + 1 )
\begin{aligned}
&\left(a+\frac{1}{b}\right)^{2}+\left(b+\frac{1}{c}\right)^{2}+\left(c+\frac{1}{a}\right)^{2} \geq \frac{\left(a+\frac{1}{b}+b+\frac{1}{c}+c+\frac{1}{a}\right)^{2}}{3} \geq \frac{(a+b+c+3)^{2}}{3}= \\
&=\frac{(a+b+c)(a+b+c)+6(a+b+c)+9}{3} \geq \frac{(a+b+c) 3 \sqrt[3]{a b c}+6(a+b+c)+9}{3}= \\
&=\frac{9(a+b+c)+9}{3}=3(a+b+c+1)
\end{aligned}
( a + b 1 ) 2 + ( b + c 1 ) 2 + ( c + a 1 ) 2 ≥ 3 ( a + b 1 + b + c 1 + c + a 1 ) 2 ≥ 3 ( a + b + c + 3 ) 2 = = 3 ( a + b + c ) ( a + b + c ) + 6 ( a + b + c ) + 9 ≥ 3 ( a + b + c ) 3 3 ab c + 6 ( a + b + c ) + 9 = = 3 9 ( a + b + c ) + 9 = 3 ( a + b + c + 1 )
The equality holds if and only if a = b = c = 1 a=b=c=1 a = b = c = 1