If $ab+bc+ca\ge1$, prove that $\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}\ge\frac{\sqrt{3}}{abc}$

Question

The following problem is from CHKMO 2018 Problem 1:

If $ab+bc+ca\ge1$, prove that $$\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}\ge\frac{\sqrt{3}}{abc}$$

I tried to use Cauchy–Schwarz inequality, by try multiplying different things, such as $1^2+1^2+1^2$, $(abc)^2+(abc)^2+(abc)^2$. But I still can’t solve it. Can someone help me?

Answer 1

Let $x=\dfrac{1}{a}, y=\dfrac{1}{b}, z=\dfrac{1}{c}$. It is easy to get $$x^2+y^2+z^2\ge xy+yz+zx$$ which is $$\dfrac{1}{a^2}+\dfrac{1}{b^2}+\dfrac{1}{c^2}\ge\dfrac{1}{ab}+\dfrac{1}{bc}+\dfrac{1}{ca}=\dfrac{a+b+c}{abc}$$ We try to prove $a+b+c\ge\sqrt{3}$. As $a^2+b^2+c^2\ge ab+bc+ca\ge1$, $$(a+b+c)^2=a^2+b^2+c^2+2ab+2bc+2ca\ge3\\ a+b+c\ge\sqrt{3}$$