-2
$\begingroup$

Let $a,b,c$ be positive real numbers such that $abc=1$. Prove that $$ {\left(a+\frac{1}{b}\right)^2}+{\left(b+\frac{1}{c}\right)^2} +{\left(c+\frac{1}{a}\right)^2}≥3(a+b+c+1)$$

My solution:

By Jensen's inequality applied to the convex function $f(x)=x^2$, for $x_1=a+\frac{1}{b} , x_2=b+\frac{1}{c}, x_3=c+\frac{1}{a}$: $$f\left(\frac{\left(a+\frac{1}{b}\right)+\left(b+\frac{1}{c}\right)+\left(c+\frac{1}{a}\right)}{3}\right)≤ \frac{{\left(a+\frac{1}{b}\right)^2}+{\left(b+\frac{1}{c}\right)^2} +{\left(c+\frac{1}{a}\right)^2}}{3}$$ $$\frac{\left(a^2bc+ab^2c+abc^2+bc+ac+ab\right)^2}{a^2b^2c^2}/9≤\frac{{\left(a+\frac{1}{b}\right)^2}+{\left(b+\frac{1}{c}\right)^2} +{\left(c+\frac{1}{a}\right)^2}}{3}$$ If we use the condition $abc=1$: $$\frac{\left(a+b+c+ab+bc+ca\right)^2}{3}≤{\left(a+\frac{1}{b}\right)^2}+{\left(b+\frac{1}{c}\right)^2} +{\left(c+\frac{1}{a}\right)^2}$$ Then we have prove that: $$\frac{\left(a+b+c+ab+bc+ca\right)^2}{3}≥3(a+b+c+1)$$ $$\left(a+b+c+ab+bc+ca\right)^2≥ 9(a+b+c+1)$$ Notice that by A.M-G.M inequality we have $ab+bc+ca≥3$: $$(a+b+c+3)^2≥9(a+b+c+1)$$ If we say $x$ to $a+b+c$ : $$(x+3)^2≥9(x+1)$$ $$x^2+6x+9≥9x+9$$ $$x^2≥3x$$ $$x≥3$$ $$a+b+c≥3$$ This statement is true by A.M-G.M inequality.

If there is another solution, I would be grateful if you could show me :)

$\endgroup$
5
  • 1
    $\begingroup$ On AoPS: artofproblemsolving.com/community/c6h1793628p11878614 $\endgroup$
    – Martin R
    Jan 21 at 8:59
  • $\begingroup$ I opened this topic because solutions in this topic are different than solutions in the linked topic. We need to merge these topics before to close. $\endgroup$
    – Michael Rozenberg
    Jan 21 at 16:20
  • 1
    $\begingroup$ @MichaelRozenberg: This has been asked and answered before (math.stackexchange.com/q/2834933/42969, math.stackexchange.com/q/3803211/42969), it is a clear duplicate. I don't understand why you reopened the question. That is not necessary for merging. $\endgroup$
    – Martin R
    Jan 21 at 16:23
  • $\begingroup$ @Bill Dubuque See please better: The topic-starter are looking for another way. $\endgroup$
    – Michael Rozenberg
    Jan 21 at 16:44
  • $\begingroup$ @ Bill Dubuque @ Michael Rozenberg I apologize for causing confusion. But when I shared this post, I didn't know it was shared. $\endgroup$
    – Rehman
    Jan 21 at 18:18

3 Answers 3

Reset to default
1
$\begingroup$

Using the inequality:

$$x^2+y^2+z^2 \geq xy+yz+zx,$$

we have

$\left(a+\frac{1}{b} \right)^2+\left(b+\frac{1}{c} \right)^2+\left(c+\frac{1}{a} \right)^2$

$ \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)$

$=\left( ab+1+\frac{a}{c}+\frac{1}{bc} \right)+\left( bc+1+\frac{b}{a}+\frac{1}{ca} \right)+\left( ca+1+\frac{c}{b}+\frac{1}{ab} \right)$

$=\left(ab+\frac{b}{a}\right)+\left(bc+\frac{c}{b}\right)+\left(ca+\frac{a}{c}\right)+\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ca}+3$

$\geq 2 \sqrt{ab\cdot\frac{b}{a}}+2 \sqrt{bc \cdot \frac{c}{b}}+2 \sqrt{ca \cdot \frac{a}{c}}+c+a+b+3$

$= 2b+2c+2a+a+b+c+3$

$=3(a+b+c+1)$

$\endgroup$
1
$\begingroup$

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 by AM-GM we obtain: $$\sum_{cyc}\left(a+\frac{1}{b}\right)^2-3(a+b+c+1)=\sum_{cyc}\left(\frac{(x+z)^2}{y^2}-\frac{3x}{y}-1\right)=$$ $$=\sum_{cyc}\left(\frac{x^2}{y^2}+1-\frac{2x}{y}+\frac{1}{2}\left(\frac{x^2}{z^2}+\frac{z^2}{y^2}-\frac{2x}{y}\right)+\frac{2xz}{y^2}-2\right)\geq0.$$

$\endgroup$
1
$\begingroup$

Another way.

By AM-GM $$\sum_{cyc}\left(a+\frac{1}{b}\right)^2-3(a+b+c+1)=\sum_{cyc}\left(a^2+\frac{1}{a^2}-3a-1+\frac{2a}{b}\right)\geq$$ $$\geq\sum_{cyc}\left(a^2+\frac{1}{a^2}-3a+1\right)=\sum_{cyc}\left(a^2+\frac{1}{a^2}-3a+1+3\ln{a}\right)\geq0.$$

$\endgroup$
1

Not the answer you're looking for? Browse other questions tagged .