Prove $a^2+b^2+c^2\gt \frac {1}{2018}$ given $\left({3a + 28b + 35c}\right)\left({20a + 23b +33c}\right) = 1$ Let $a, b, c$ be real numbers such that $\left({3a + 28b + 35c}\right)\left({20a + 23b +33c}\right) = 1$. Prove that $a^2+b^2+c^2\gt \frac {1}{2018}$.
It looks like an easy question, but I thought for a while and could not figure it out. Can anyone give me some hints? Thank you. 

By C-S $$a^2+b^2+c^2=$$
$$=\frac{\sqrt{(3^2+28^2+35^2)(a^2+b^2+c^2)\cdot(20^2+23^2+33^2)(a^2+b^2+c^2)}}{2018}\geq$$
$$\geq\frac{(3a+28b+35c)(20a+23b+33c)}{2018}=\frac{1}{2018}.$$
