If $a, b, c$ are positive numbers with $abc=1$, then $(a+\frac{1}{b})^2+(b+\frac{1}{c})^2+(c+\frac{1}{a})^2\ge 3(a+b+c+1)$.

Question

I was just doing an inequality question from a past international competition, but:

a/ I'm not completely certain about my proof

b/ I wonder if there is a simpler and less laborious method of solving it

The question is as follow:

Let $a, b, c$ be positive real numbers such that $abc=1$. Prove that:

$$(a+\frac{1}{b})^2+(b+\frac{1}{c})^2+(c+\frac{1}{a})^2\ge 3(a+b+c+1).$$

I solved it as follows:

$$\begin{aligned}&\quad(a+\frac{1}{b})^2+(b+\frac{1}{c})^2+(c+\frac{1}{a})^2\\ &=a^2+\frac{2a}{b}+\frac{1}{b^2}+b^2+\frac{2b}{c}+\frac{1}{c^2}+c^2+\frac{2c}{a}+\frac{1}{a^2}\\ &=a^2+\frac{a}{b}+\frac{1}{b^2}+b^2+\frac{b}{c}+\frac{1}{c^2}+c^2+\frac{c}{a}+\frac{1}{a^2}+(\frac{a}{b}+\frac{b}{c}+\frac{c}{a})\\ &\ge a^2+\frac{a}{b}+\frac{1}{b^2}+b^2+\frac{b}{c}+\frac{1}{c^2}+c^2+\frac{c}{a}+\frac{1}{a^2}+3\quad \text{(AM-GM)}\\ &=a^2+\frac{a}{b}+\frac{ac}{b}+b^2+\frac{b}{c}+\frac{ab}{c}+c^2+\frac{c}{a}+\frac{bc}{a}+3\\ &=a^2+b^2+c^2+\frac{a}{b}+\frac{b}{c}+\frac{c}{a}+\frac{ac}{b}+\frac{bc}{a}+\frac{ab}{c}+3\\ &=(a^2+\frac{b}{c}+\frac{ac}{b})+(b^2+\frac{c}{a}+\frac{ba}{c})+(c^2+\frac{a}{b}+\frac{bc}{a})\\ &\ge 3\sqrt[3]{a^3}+3\sqrt[3]{b^3}+3\sqrt[3]{c^3}+3\quad\text{(AM-GM)}\\ &=3(a+b+c+1) \end{aligned}$$

Could you please tell me if my solution is correct and also show me an easier and shorter approach to the question?

Answer 1

Your proof is correct, but by using a cyclic sum it looks much more shorter:

By AM-GM $$\sum_{cyc}\left(a+\frac{1}{b}\right)^2=\sum_{cyc}\left(a^2+\frac{2a}{b}+\frac{1}{b^2}\right)=\sum_{cyc}\left(a^2+\frac{b}{c}+\frac{ac}{b}+\frac{a}{b}\right)\geq$$ $$\geq\sum_{cyc}\left(3\sqrt{a^2\cdot\frac{b}{c}\cdot\frac{ac}{b}}+\sqrt[3]{\prod_{cyc}\frac{a}{b}}\right)=3(a+b+c+1).$$

Another way:

Let $a=\frac{x}{y}$ and $b=\frac{y}{z},$ where $x$, $y$ and $z$ are positives.

Thus, $c=\frac{z}{x}$ and we need to prove that: $$\sum_{cyc}\left(\frac{x}{y}+\frac{z}{y}\right)^2\geq\sum_{cyc}\left(\frac{3x}{y}+1\right)$$ or $$\sum_{cyc}(x^4y^2+x^4z^2+2x^3y^3-3x^3z^2y-x^2y^2z^2)\geq0,$$ which is true by AM-GM and Rearrangement: $$\sum_{cyc}(x^4y^2+x^4z^2+2x^3y^3-3x^3z^2y-x^2y^2z^2)\geq$$ $$\geq\sum_{cyc}(2x^3y^3+x^3y^3+x^2y^2z^2-3x^3z^2y-x^2y^2z^2)=$$ $$=3\sum_{cyc}((xy)^3-(xz)^2(xy))\geq0.$$

Answer 2

I don't know if my solution is more simpler:
Suppose $a={x\over y} $, $b = {y\over z} $ and $c = {z\over x} $. (just to eliminate the condition $abc=1$) And then expanding, $$ \sum_{cyc}{x^2\over y^2}+\sum_{cyc}{y^2\over x^2}+2\sum_{cyc}{yz\over x^2} \geq 3\left( \sum_{cyc}{x\over y}+1\right) $$ $$ \Rightarrow 2\sum_{cyc}{x\over y}+\sum_{cyc}{y^2\over x^2}+2\sum_{cyc}{yz\over x^2} \geq 3\sum_{cyc}{x\over y} + 6 $$ $$ \Rightarrow \sum_{cyc}{y^2\over x^2} +2\sum_{cyc}{yz\over x^2} \geq \sum_{cyc}{x\over y} + 6 $$ Notice that $$ {\sum_{cyc}{y^2\over x^2} + \sum_{cyc}{y^2\over x^2} \over 2} \ge \sum_{cyc}{x\over y} $$ Therefore, it remains to prove that: $$ \sum_{cyc}{yz\over x^2} \ge 3 $$ Which is simple AM-GM. Done!