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

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!

• What does MO stand for? Do you have a reference for the "England MO" inequality? – saulspatz Apr 3 '18 at 13:36

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 $$(a-b)\left(c(a-b)+a^2+a(a+c-b)\right)+b(a-c)(b-c)\geq 0$$
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.