By applying $x^2+y^2+z^2 \ge xy+yz+zx$ for all $x,y,z$ we have \[\sum \left( a+ \frac{1}{b} \right)^2 \ge \sum \left( a+ \frac 1b \right) \left(b+ \frac 1c \right)=\sum ab+ \sum \frac ac+ \sum \frac{1}{ab}+3.\] Since $abc=1$ so $\sum \frac{1}{ab}=\sum a$. Hence, we need to prove $\sum ab+ \sum \frac ac \ge 2 \sum a. \qquad (1)$ Using AM-GM inequality we have \[ab+ \frac{b}{a} \ge 2b, bc+ \frac{c}{b} \ge 2c, ca+ \frac{a}{c} \ge 2a.\] Thus $(1)$ is true. The equality holds if and only if $a=b=c=1$. $\blacksquare$

hello, plugging $a=\frac{x}{y},b=\frac{y}{z},c=\frac{x}{z}$ we get ${x}^{4}{y}^{2}+{x}^{4}{z}^{2}+2\,{x}^{3}{y}^{3}-3\,{x}^{3}y{z}^{2}+2\, {x}^{3}{z}^{3}+{x}^{2}{y}^{4}-3\,{y}^{3}z{x}^{2}-3\,{y}^{2}{z}^{2}{x}^ {2}+{x}^{2}{z}^{4}-3\,{z}^{3}{y}^{2}x+{y}^{4}{z}^{2}+2\,{y}^{3}{z}^{3} +{y}^{2}{z}^{4} \geq 0$ and with $y=x+u,z=x+u+v$ we obtain $ \left( 11\,{u}^{2}+11\,uv+11\,{v}^{2} \right) {x}^{4}+ \left( 35\,{u} ^{3}+51\,{u}^{2}v+34\,u{v}^{2}+9\,{v}^{3} \right) {x}^{3}+ \left( 41\, {u}^{4}+79\,{u}^{3}v+54\,{u}^{2}{v}^{2}+16\,u{v}^{3}+2\,{v}^{4} \right) {x}^{2}+ \left( 21\,{u}^{5}+51\,{u}^{4}v+43\,{u}^{3}{v}^{2}+ 15\,{u}^{2}{v}^{3}+2\,u{v}^{4} \right) x+4\,{u}^{6}+12\,{u}^{5}v+13\,{ u}^{4}{v}^{2}+6\,{u}^{3}{v}^{3}+{u}^{2}{v}^{4} \geq 0$ Sonnhard.

We can transform $a+1/b$ to $a+ac=a(c+1)$.We do that for every one of them and then just write it down.We get that our inequality is equivalent to: $a^2 + b^2 + c^2 + a^2 b^2 + a^2 c^2 + b^2 c^2 + 2ab^2 + 2bc^2 + 2ca^2 >= 3a + 3b + 3c + 3$.By aplying $AM-GM$ we have that: $a^2 + a^2 c^2 +ab^2 >= 3a$ ( we do similar for other to ).Also we get by $AM-GM$ that: $ab^2+bc^2+ca^2>=3$ so the inequailiy is proven.

We have \[ a^{2}+\frac{1}{b^{2}}+\frac{b}{c}\geq3\sqrt[3]{a^{2}\cdot\frac{1}{b^{2}}\cdot\frac{b}{c}}=3\sqrt[3]{\frac{a^{2}}{bc}}=3\sqrt[3]{\frac{a^{2}\cdot abc}{bc}}=3a. \] In analogous way we have \[ b^{2}+\frac{1}{c^{2}}+\frac{c}{a}\geq3b, \] \[ c^{2}+\frac{1}{a^{2}}+\frac{a}{b}\geq3c. \] Also \[ \frac{a}{b}+\frac{b}{c}+\frac{c}{a}\geq3\sqrt[3]{\frac{a}{b}\cdot\frac{b}{c}\cdot\frac{c}{a}}=3. \] Adding up these inequalities we are done. Equality is achieved if and only if $a=b=c=1.$

From CBS: we have that $LHS \ge \dfrac{(a+b+c+\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c})^2}{3} \ge \dfrac{(a+b+c+3)^2}{3}\ge 3(a+b+c+1)<=>a+b+c \ge 3$ truly.I used $\dfrac{1}{b}+\dfrac{1}{c} \ge 3$(AM-GM) and

For positive real numbers $a,b,c$ with $abc\leq1$ prove that $\left(a+\frac{1}{b}\right)^{2}+\left(b+\frac{1}{c}\right)^{2}+\left(c+\frac{1}{a}\right)^{2}\geq 3(a+b+c+1)$ Proof:$\left(a+\frac{1}{b}\right)^{2}+\left(b+\frac{1}{c}\right)^{2}+\left(c+\frac{1}{a}\right)^{2}$ $=\sum \left((a+\frac{1}{b})^{2}+4\right)-12\geq 4\sum \left(a+\frac{1}{b}\right)-12$ $=4\sum a+4\sum \frac{1}{a}-12\geq 3\sum a +3\sqrt[3]{ abc}+12\sqrt[3]{\frac{1}{abc}}-12$ $\geq 3(a+b+c+1).$ The following inequality is also true. For positive real numbers $a,b,c,d$ with $abcd\leq 1$ prove that \[\left(a+\frac{1}{b}\right)^{2}+\left(b+\frac{1}{c}\right)^{2}+\left(c+\frac{1}{d}\right)+\left(d+\frac{1}{a}\right)^{2}\geq 3(a+b+c+d)+4.\]

By AM-GM, $f(a):=a+\frac{4}{a}+3\left(1-\frac{1}{a}\right)\geq 5$ etc. we have $f(a)+f(b)+f(c)$ $\Longleftrightarrow a+b+c+4\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)\geq 3\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)+6\geq 15\cdots (*)$ by AM-GM Since $x^2\geq 4x-4$, we have $\left(a+\frac{1}{b}\right)^2+\left(b+\frac{1}{c}\right)^2+\left(c+\frac{1}{a}\right)^2$ $\geq 4(a+b+c)+4\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)-12 \geq 3(a+b+c+1)\ \because (*)$

Applying Cauchy-Shwartz we get: $((a+\frac{1}{b})^2+(b+\frac{1}{c})^2+(c+\frac{1}{a})^2)(1^2+1^2+1^2)\geq(a+b+c+\frac{1}{a}+\frac{1}{b}+\frac{1}{c})^2 \geq(a+b+c+3)^2$ Using the fact $abc=1$. Then we have: $(a+\frac{1}{b})^2+(b+\frac{1}{c})^2+(c+\frac{1}{a}\geq \frac{(a+b+c+3)^2}{3}\geq 3\cdot (a+b+c+1)$,which is trivially true by AM-GM since $a^2 + ab+ac\geq 3a$. $Q.E.D.$

sqing wrote:

For positive real numbers $a,b,c,d$ with $abcd\leq 1$ prove that \[\left(a+\frac{1}{b}\right)^{2}+\left(b+\frac{1}{c}\right)^{2}+\left(c+\frac{1}{d}\right)+\left(d+\frac{1}{a}\right)^{2}\geq 3(a+b+c+d)+4.\] $\left(a+\frac{1}{b}\right)^{2}+\left(b+\frac{1}{c}\right)^{2}+\left(c+\frac{1}{d}\right)+\left(d+\frac{1}{a}\right)^{2}$ $=\sum \left((a+\frac{1}{b})^{2}+4\right)-16\geq 4\sum \left(a+\frac{1}{b}\right)-16$ $=4\sum a+4\sum \frac{1}{a}-16\geq 3\sum a +4\sqrt[4]{ abcd}+16\sqrt[4]{\frac{1}{abcd}}-16$ $\geq 3\sum a +12\sqrt[4]{\frac{1}{abcd}}-8\geq 3(a+b+c+d)+4.$ here here For positive real numbers $a_1, a_2, ..., a_n$ with $a_1a_2...a_n \leq 1$ prove that \[\left(a_1 + \dfrac{1}{a_2}\right)^2 + \left(a_2 + \dfrac{1}{a_3}\right)^2 + ... + \left(a_n + \dfrac{1}{a_1}\right)^2 \ge 3(a_1 + a_2 + ... + a_n) + n.\]

A more stronger inequality is: For positive real numbers $a,b,c$ with $abc=1$ prove that \[\left(a+\frac{1}{b}\right)^{2}+\left(b+\frac{1}{c}\right)^{2}+\left(c+\frac{1}{a}\right)^{2}\geq 6(a+b+c-1)\]

MariusStanean wrote: A more stronger inequality is: For positive real numbers $a,b,c$ with $abc=1$ prove that \[\left(a+\frac{1}{b}\right)^{2}+\left(b+\frac{1}{c}\right)^{2}+\left(c+\frac{1}{a}\right)^{2}\geq 6(a+b+c-1)\] See here http://www.artofproblemsolving.com/Forum/viewtopic.php?t=595138&p3530990#p3530990

$\left(a+\frac{1}{b}\right)^2+\left(b+\frac{1}{c}\right)^2+\left(c+\frac{1}{a}\right)^2$ $\geq 4\left(a+\frac{1}{b}\right)+4\left(b+\frac{1}{c}\right)+4\left(c+\frac{1}{a}\right)-12$ $=3(a+b+c)+a+b+c+\frac{4}{a}+\frac{4}{b}+\frac{4}{c}-12$ $\geq 3(a+b+c)+15\sqrt[15]{a\cdot b\cdot c\cdot \left(\frac{1}{a}\right)^4\left(\frac{1}{b}\right)^4\left(\frac{1}{c}\right)^4}-12$ $=3(a+b+c+1)\ Q.E.D.$

$\left(a+\frac 1b\right)^2 + \left(b+ \frac 1c\right)^2 + \left(c+ \frac 1a\right)^2$ $\geq$ $\frac{\left(a+b+c+\frac 1a+\frac 1b+\frac 1c\right)^2}{3}$ $\geq$ $\frac{\left(a+b+c+3\right)^2}{3}$ $=$ $\frac{\left(s+3\right)^2}{3}$. We have to prove that $\frac{\left(s+3\right)^2}{3}$ $\geq$ $3(s+1)$ $\Leftrightarrow$ $s^2+6s+9$ $\geq $ $9s + 9$ $\Leftrightarrow$ $s^2 \geq 3s$ $\Leftrightarrow$ $s \geq 3$, which is obvious, by $AM \geq GM$. The equality holds when $a=b=c=1$ (in $AM \geq GM$ and $QM \geq AM$)

SO MANY SOLUTIONS FOR ONE PROBLEM

$3(a^2+b^2+c^2) \geq (a+b+c)^2 \geq 3(a+b+c)$ [by CS and AM-GM] $\sum{\frac{a}{b}} = \sum{ab^2} \geq a+b+c $ [ as $2a^2c+ab^2 \geq 3a $] [by AM-GM] $\sum{\frac{1}{a^2} \geq 3}$ .[by AM-GM]. Adding we are done.

gavrilos wrote: For positive real numbers $a,b,c$ with $abc=1$ prove that $\left(a+\frac{1}{b}\right)^{2}+\left(b+\frac{1}{c}\right)^{2}+\left(c+\frac{1}{a}\right)^{2}\geq 3(a+b+c+1)$ Usual fare: Letting $a=x/y$, $b=y/z$, $c=z/x$ the inequality in question becomes \[ \sum_{\text{cyc}} \frac{(x+y)^2}{z^2} \ge 3 \left( \frac xy + \frac yz + \frac zx + 1 \right). \] Clearing all denominators, it suffices to prove that \[ \sum_{\text{cyc}} x^2y^2(x+y)^2 \ge 3xyz \left( x^2z+y^2x+z^2y \right) + 3(xyz)^2 \] which is quite weak. You can easily show (or use Muirhead to show) that the left-hand side is at least $4(x^3y^3+y^3z^3+z^3x^3)$. But, $3xyz \cdot x^2z \le x^3y^3 + 2x^3z^3$ and $3(xyz)^2 \le x^3y^3+y^3z^3+z^3x^3$ by AM-GM, hence, done.

How do u guys come up with these solutions?

gavrilos wrote: For positive real numbers $a,b,c$ with $abc=1$ prove that $\left(a+\frac{1}{b}\right)^{2}+\left(b+\frac{1}{c}\right)^{2}+\left(c+\frac{1}{a}\right)^{2}\geq 3(a+b+c+1)$ By computer,we have $$LHS-RHS=\frac{1}{2}\sum_{cyc}{(a-b)^2}+\sum_{cyc}{\frac{(2ac+a+1)(a-1)^2}{a^2}}\ge{0}$$

After plugging a=x/y ,b=y/z,c=z/x can we use rearrangement assuming x>y>z???

gavrilos wrote: For positive real numbers $a,b,c$ with $abc=1$ prove that $\left(a+\frac{1}{b}\right)^{2}+\left(b+\frac{1}{c}\right)^{2}+\left(c+\frac{1}{a}\right)^{2}\geq 3(a+b+c+1)$ The following inequality is also true. For positive real numbers $a,b,c$ with $abc=1$ prove that $$\left(a+\frac{1}{b}\right)^{2}+\left(b+\frac{1}{c}\right)^{2}+\left(c+\frac{1}{a}\right)^{2}+4\geq (a+b+c+1)^2$$

hey check my question above can we use rearrangement inequality(assuming x>Y>Z) after substituting a=x/y,b=y/z,c=z/x?????????????

The LHS is equivalent to $$\sum \left(a^2 + \frac{2a}{b} + \frac{1}{b^2}\right)$$We can cycle the middle term (it is a cyclic sum) to get: $$\sum \left(a^2 + \frac{2b}{c} + \frac{1}{b^2}\right)$$Substitute $abc = 1$ $$\sum\left(a^2 + 2ab^2 + \frac{1}{b^2}\right)$$$a^2 + ab^2 + \frac{1}{b^2} \geq 3a$ by AM-GM and $\sum ab^2 = ab^2 + bc^2 + ca^2 \geq 3abc = 3$ so therefore $$\sum \left(a^2 + ab^2 + \frac{1}{b^2}\right) + \sum ab^2 \geq \sum 3a + 3$$$$\sum\left(a^2 + 2ab^2 + \frac{1}{b^2}\right) \geq \sum3a + 3$$Done!

Here's another one! Solution: Let $a^{\frac {1}{3}}=x, b^{\frac {1}{3}}=y $, and $c^{\frac {1}{3}}=z $. Applying A.M.-G.M., we have the following: $x^8y^2z^2+a^2c^2+x^4y^7z\ge 3a^2bc=3a $, $x^2y^8z^2+a^2b^2+xy^4z^7\ge 3ab^2c=3b $, $x^2y^2z^8+b^2c^2+x^7yz^4\ge 3abc^2=3c $, $x^7yz^4+x^4y^7z+xy^4z^7\ge 3$. Adding the above inequalities, we get, $\sum_{\text {cyc}}x^8y^2z^2+\sum_{\text {cyc}}a^2b^2+2\sum_{\text{cyc}}x^7yz^4\ge 3\sum_{\text {cyc}}a+3$, which is equivalent to the given inequality as $abc=1$.

$(a+\frac{1}{b})^{2}+(b+\frac{1}{c})^{2}+(c+\frac{1}{a})^{2} =\sum_{cyc}a^{2}+\sum_{cyc}\frac{a}{b}+\sum_{cyc}(\frac{1}{a})^{2} = S$ Now by AM-GM we get: $S\geq \sum_{cyc}a^{2}+6+\sum_{cyc}a$ So we should prove that: $\sum_{cyc}a^{2}+6+\sum_{cyc}a\geq 3\sum_{cyc}a +3\Rightarrow a^{2}+b^{2}+c^{2}\geq 2(a+b+c)-3$ Which can be rewritten as: $(a-1)^{2}+(b-1)^{2}+(c-1)^{2}\geq0$ Which is obviously true.

gavrilos wrote: For positive real numbers $a,b,c$ with $abc=1$ prove that $\left(a+\frac{1}{b}\right)^{2}+\left(b+\frac{1}{c}\right)^{2}+\left(c+\frac{1}{a}\right)^{2}\geq 3(a+b+c+1)$ Solution: We would use the famous Ravi substitution : $a=\dfrac {x}{y},b=\dfrac{y}{z}$ and$ c$=$\dfrac{z}{x}$ Hence the given inequality transforms into: $\sum \dfrac {(x+z)^2}{y^2} \ge 3(\dfrac{\sum x^2z}{xyz})$ $\implies \sum(x+z)^2 x^2 z^2 \ge 3\sum x^3yz^2 + 3x^2y^2z^2$ $\implies \sum x^4z^2 +2x^3z^3 \ge 3\sum x^3yz^2 + 3(xyz)^2$ And this is a Weak form of Muirhead's inequality as: $(3,3,0)$ majorizes $ (3,1,2)$ And:$(4,0,2)$ majorizes $ (2,2,2)$ $QED$

Wait, gavrilos wrote: For positive real numbers $a,b,c$ with $abc=1$ prove that $\left(a+\frac{1}{b}\right)^{2}+\left(b+\frac{1}{c}\right)^{2}+\left(c+\frac{1}{a}\right)^{2}\geq 3(a+b+c+1)$ So $a, b, c$ are positive real, and $abc = 1$ It's automatically gives that $a=b=c=1$

AbsoluSquare wrote: Wait, gavrilos wrote: For positive real numbers $a,b,c$ with $abc=1$ prove that $\left(a+\frac{1}{b}\right)^{2}+\left(b+\frac{1}{c}\right)^{2}+\left(c+\frac{1}{a}\right)^{2}\geq 3(a+b+c+1)$ So $a, b, c$ are positive real, and $abc = 1$ It's automatically gives that $a=b=c=1$ Oh sorry it's wrong, taught $a, b, c$ were integers

Satops wrote: gavrilos wrote: For positive real numbers $a,b,c$ with $abc=1$ prove that $\left(a+\frac{1}{b}\right)^{2}+\left(b+\frac{1}{c}\right)^{2}+\left(c+\frac{1}{a}\right)^{2}\geq 3(a+b+c+1)$ Solution: We would use the famous Ravi substitution : $a=\dfrac {x}{y},b=\dfrac{y}{z}$ and$ c$=$\dfrac{z}{x}$ Hence the given inequality transforms into: $\sum \dfrac {(x+z)^2}{y^2} \ge 3(\dfrac{\sum x^2z}{xyz})$ $\implies \sum(x+z)^2 x^2 z^2 \ge 3\sum x^3yz^2 + 3x^2y^2z^2$ $\implies \sum x^4z^2 +2x^3z^3 \ge 3\sum x^3yz^2 + 3(xyz)^2$ And this is a Weak form of Muirhead's inequality as: $(3,3,0)$ majorizes $ (3,1,2)$ And:$(4,0,2)$ majorizes $ (2,2,2)$ $QED$ Umm I don't think $(4, 0, 2)$ majorizes $(2, 2, 2)$ as majorization requires weakly decreasing sequences. But I think you can just reorder '$x$, $y$ and $z$' to '$x$, $z$ and $y$' to solve this issue.

Similarly, by $AM-QM$ inequality to get that $(a+\frac{1}{b})^2+(b+\frac{1}{c})^2+(c+\frac{1}{a})^2 \geq \frac{(a+\frac{1}{b}+b+\frac{1}{c}+c+\frac{1}{a})}{3}$ From $AM-GM$ we get that $\frac{1}{a}+\frac{1}{b}+\frac{1}{c} \geq \sqrt[3]{\frac{1}{abc}}=3$. So, $\frac{(a+\frac{1}{b}+b+\frac{1}{c}+c+\frac{1}{a})}{3} \geq \frac{(a+b+c+3)^2}{3}=\frac{(a+b+c)(a+b+c)+6(a+b+c)+9}{3}=X$.Again, by $AM-GM$ $X \geq \frac{3(a+b+c)\sqrt[3]{abc}+6(a+b+c)+9}{3}=\frac{9(a+b+c+1)}{3}=3(a+b+c+1)$. And that is all

For positive real numbers $a,b,c$ with $abc=1$ prove that \[\left(a+\frac{1}{b}\right)^{3}+\left(b+\frac{1}{c}\right)^{3}+\left(c+\frac{1}{a}\right)^{3}\geq 12(a^{2}+b^{2}+c^{2}-1)\]For positive real numbers $a,b,c$ with $ab+bc+ca=3$ prove that \[\left(a+\frac{1}{b}\right)\left(b+\frac{1}{c}\right)\left(c+\frac{1}{a}\right)\geq 2\sqrt{2(a+b+c-1)}\] MariusStanean wrote: A more stronger inequality is: For positive real numbers $a,b,c$ with $abc=1$ prove that \[\left(a+\frac{1}{b}\right)^{2}+\left(b+\frac{1}{c}\right)^{2}+\left(c+\frac{1}{a}\right)^{2}\geq 6(a+b+c-1)\]

$\left(a+\frac{1}{b}\right)^{2}+\left(b+\frac{1}{c}\right)^{2}+\left(c+\frac{1}{a}\right)^{2}=(a^2+b^2+c^2)+(\frac{a}{b}+\frac{b}{c}+\frac{c}{a})+(\frac{1}{b^2}+\frac{1}{c^2}+\frac{1}{a^2})+(\frac{a}{b}+\frac{b}{c}+\frac{c}{a})\geq (\dfrac{(a+b+c)^2}{3})+(\frac{a}{b}+\frac{b}{c}+\frac{c}{a})+(\frac{1}{bc}+\frac{1}{ac}+\frac{1}{ab})+3 \geq (a+b+c)+(\frac{a}{b}+\frac{b}{c}+\frac{c}{a})+(a+b+c)+3$ Then it is sufficient to prove that $\frac{a}{b}+\frac{b}{c}+\frac{c}{a} \geq a+b+c$ Using $abc=1$ we have $c=\frac{1}{ab}$ so we should prove that $(\frac{a}{b}+b^2a+\frac{1}{a^2b}) \geq a+b+\frac{1}{ab}$ but by AM-GM we have $$\frac{\frac{a}{b}+\frac{a}{b}+b^2a}{3} \geq a$$$$\frac{b^2a+b^2a+\frac{1}{a^2b}}{3} \geq b$$$$\frac{\frac{1}{a^2b}+\frac{1}{a^2b}+\frac{a}{b}}{3} \geq \frac{1}{ab}$$ By adding them all, we came to the conclusion. The equality holds if and only if $a=b=c=1$ $\blacksquare$

$AM-GM$ alone is sufficient to destroy the problem

Cute and intuitive. We have that $\sum_{cyc}(a+\frac 1 b)^2\geq \sum_{cyc} ab+\frac{1}{ab}+\frac{b}{a}+1 = \sum_{cyc} ab+a+\frac{b}{a}+1,$ and we want to show that this is greater than $3(a+b+c)+3,$ which amounts to proving that $\sum_{cyc} ab+\frac{b}{a}\geq 2(a+b+c)$. But this is trivial by AM-GM, so we are done.

Generalization 1 For positive reel numbers $a,b,c$ with $abc=k^3$ prove that $$\left(a+\dfrac{k}{b}\right)^2+\left(b+\dfrac{k}{c}\right)^2+\left(c+\dfrac{k}{a}\right)^2\geq 3(a+b+c+1)$$

Generalization 2 For positive reel numbers $a_{1},a_{2},\cdots,a_{n}$ ($n>2$) with $\prod{a_{1}}=\left(n-2\right)^n$ . Then prove that $$\sum_{cyc}{\left(a_{1}+\dfrac{n-2}{a_{2}}\right)^2}\geq n\left(\sum_{cyc}{a_{1}} +1\right)$$

Generalized JBMO 2014 #3

$\left(a+\frac{1}{b}\right)^{2}+\left(b+\frac{1}{c}\right)^{2}+\left(c+\frac{1}{a}\right)^{2}=\sum \left( a+ \frac{1}{b} \right)^2=\sum \left( a^2+ \frac{2a}{b} +\frac{1}{b^2} \right)=a^2+b^2+c^2+ \frac{2a}{b} + \frac{2b}{c} + \frac{2a}{c}+ \frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}$ $\left(a+\frac{1}{b}\right)^{2}+\left(b+\frac{1}{c}\right)^{2}+\left(c+\frac{1}{a}\right)^{2} =a^2+b^2+c^2+ \frac{2a}{b} + \frac{2b}{c} + \frac{2a}{c}+ \frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}$ From the condition we have: $abc=1$ $\implies$ $c=\frac{1}{ab}$ So $\left(a+\frac{1}{b}\right)^{2}+\left(b+\frac{1}{c}\right)^{2}+\left(c+\frac{1}{a}\right)^{2} =a^2+b^2+c^2+ \frac{2a}{b} + \frac{2b}{c} + \frac{2a}{c}+ \frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}=a^2+b^2+\frac{1}{a^2b^2}+\frac{2a}{b}+2ab^2+\frac{2}{a^2b}+\frac{1}{a^2}+\frac{1}{b^2}+a^2b^2$ $\left(a+\frac{1}{b}\right)^{2}+\left(b+\frac{1}{c}\right)^{2}+\left(c+\frac{1}{a}\right)^{2}=a^2+b^2+\frac{1}{a^2b^2}+\frac{2a}{b}+2ab^2+\frac{2}{a^2b}+\frac{1}{a^2}+\frac{1}{b^2}+a^2b^2$ ...(1) By $AM-GM$ we have: $a^2+ab^2+\frac{1}{b^2} \ge 3\sqrt[3]{a^2*ab^2*\frac{1}{b^2}}=3\sqrt[3]{a^3}=3a $ $b^2+a^2b^2+\frac{1}{a^2b} \ge 3\sqrt[3]{b^2*a^2b^2*\frac{1}{a^2b}}=3\sqrt[3]{b^3}=3b$ $\frac{1}{a^2}+\frac{a}{b}+\frac{1}{a^2b} \ge 3\sqrt[3]{\frac{1}{a^2}*\frac{a}{b}*\frac{1}{a^2b}}=3\sqrt[3]{\frac{1}{a^3b^3}}=\sqrt[3]{c^3}=3c$ $ab^2+\frac{a}{b}+\frac{1}{a^2b} \ge 3\sqrt[3]{ab^2*\frac{a}{b}*\frac{1}{a^2b}}=3\sqrt[3]{1}=3*1=3$ So $a^2+ab^2+\frac{1}{b^2} +b^2+a^2b^2+\frac{1}{a^2b}+\frac{1}{a^2}+\frac{a}{b}+\frac{1}{a^2b}+ab^2+\frac{a}{b}+\frac{1}{a^2b} \ge 3a+3b+3c+3$ $\implies$ $ a^2+b^2+\frac{1}{a^2b^2}+\frac{2a}{b}+2ab^2+\frac{2}{a^2b}+\frac{1}{a^2}+\frac{1}{b^2}+a^2b^2 \ge 3a+3b+3c+3$ Combining with $(1)$ $\left(a+\frac{1}{b}\right)^{2}+\left(b+\frac{1}{c}\right)^{2}+\left(c+\frac{1}{a}\right)^{2}\ge 3(a+b+c+1)$ As desired.