<s>[INST] 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. 
 [/INST] By using C-S we have :
$$|3a+28b+35c| \le \sqrt{3^2+28^2+35^2} \cdot \sqrt{a^2+b^2+c^2} \tag 1$$
and
$$|20a+23b+33c|\le \sqrt{20^2+23^2+33^2} \cdot \sqrt{a^2+b^2+c^2} \tag 2$$
Multiplying $(1)$ and $(2)$
$$|(3a+28b+35c)(20+23b+33c)| \le \sqrt{20^2+23^2+33^2} \cdot \sqrt{a^2+b^2+c^2} \cdot \sqrt{3^2+28^2+35^2} \cdot \sqrt{a^2+b^2+c^2} $$
$$\implies \underbrace{(3a+28b+35c)(20+23b+33c)}_{=1} \le 2018 \cdot (a^2+b^2+c^2) $$
Thus
$$a^2+b^2+c^2 > \frac{1}{2018}$$
And, the inequality is strict because $\frac ba =\frac {28}3$ and $\frac ba =\frac{23}{20}$ cannot be simultaneously true
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