Prove $\frac{a}{b}+ \frac{b}{c}+ \frac{c}{a}\geq \frac{1}{a}+ \frac{1}{b}+ \frac{1}{c}$

Question

Give $a, b. c$ be the lengths of a triangle and $a+ b+ c= 3$. Prove that: $$\frac{a}{b}+ \frac{b}{c}+ \frac{c}{a}\geq \frac{1}{a}+ \frac{1}{b}+ \frac{1}{c}$$ My try

We have England MO inequality: $$\left ( \frac{a}{b}+ \frac{b}{c}+ \frac{c}{a} \right )^{2}\geq \left ( a+ b+ c \right )\left ( \frac{1}{a}+ \frac{1}{b}+ \frac{1}{c} \right )$$

and $$\frac{a}{b}+ \frac{b}{c}+ \frac{c}{a} \geq 3= a+ b+ c$$

But it cannot be used to prove the above inequality and how to use triangle inequality in this inequality. I need the help. Thanks!

Answer 1

After any Sheets of paper i have got $$a(b-c)(b-a)+b(a-c)(b-c)+c(a-c)(a-b)\geq 0$$ and this is $$(a-b)\left(c(a-b)+a^2+a(a+c-b)\right)+b(a-c)(b-c)\geq 0$$

Answer 2

Let $a=y+z$, $b=x+z$ and $c=x+y$.

Hence, $x$, $y$ and $z$ are positives and we need to prove that $$\sum_{cyc}y(x-y)^2\geq0,$$ which is obvious.

Answer 3

Another way.

Let $c=\min\{a,b,c\}$.

Thus, $$3\sum_{cyc}\left(\frac{a}{b}-\frac{1}{a}\right)=\sum_{cyc}\left(\frac{3a}{b}-(a+b+c)\frac{1}{a}\right)=$$ $$=\sum_{cyc}\left(\frac{2a}{b}-\frac{b}{a}-1\right)=2\left(\sum_{cyc}\frac{a}{b}-3\right)-\left(\sum_{cyc}\frac{b}{a}-3\right)=$$ $$=2\left(\frac{(a-b)^2}{ab}+\frac{(c-a)(c-b)}{ac}\right)-\left(\frac{(a-b)^2}{ab}+\frac{(c-a)(c-b)}{bc}\right)=$$ $$=\frac{(a-b)^2}{ab}+\frac{(c-a)(c-b)(2b-a)}{abc}\geq\frac{(a-b)^2}{ab}+\frac{(c-a)(c-b)(b+c-a)}{abc}\geq0.$$