# Proving an inequality given a constraint

Given $$a,b,c >0$$ and $$ab+bc+ca=3$$ prove the following inequality :

$$\large 3\left( \frac1{a} + \frac1{b} +\frac1{c} \right) \geqslant 6 + \frac{ab}{c} +\frac{bc}{a} + \frac{ca}{b}$$

My work : LHS$$= 3 \cdot \frac{ab+bc+ca}{abc} = \frac{9}{abc}\\ \geqslant 6 +\frac{ab}{c} +\frac{bc}{a} + \frac{ca}{b}$$

By AM-GM : $$ab+bc+ca=3 \geqslant 3 \sqrt[3]{(abc)^2 } \implies abc\leqslant1$$

$$\frac{9}{abc} \geqslant 9 \geqslant6+ \frac{ab}{c} +\frac{bc}{a} + \frac{ca}{b}$$

$$\frac{ab}{c} +\frac{bc}{a} + \frac{ca}{b} \leqslant 3$$

$$(ab)^2 +(bc)^2 +(ca)^2 \leqslant 3$$

Am I on the right track , if yes, how to continue further ?

We have $$3\left(\frac1a+\frac1b+\frac1c\right)-\left(6+\frac{ab}c+\frac{bc}a+\frac{ca}b\right)$$ $$=3\left(\frac1a+\frac1b+\frac1c\right)-\left(6+\frac{ab}c+\frac{bc}a+\frac{ca}b\right)+\frac{3(ab+bc+ca-3)+(ab+bc+ca-3)^2}{abc}$$ $$=2a+2b+2c-6$$ Note $$9\le\frac{(a-b)^2+(b-c)^2+(c-a)^2}2~~~~+3(ab+bc+ca)=a^2+b^2+c^2+2ab+2bc+2ca=(a+b+c)^2$$.
Therefore, $$3\le a+b+c$$ so $$2a+2b+2c-6\ge0$$ so $$3\left(\frac1a+\frac1b+\frac1c\right)\ge6+\frac{ab}c+\frac{bc}a+\frac{ca}b$$