Prove $a^2+b^2+c^2\gt \frac {1}{2018}$ given $\left({3a + 28b + 35c}\right)\left({20a + 23b +33c}\right) = 1$

Question

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.

Answer 1

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

Answer 2

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

Answer 3

With a bit of linear algebra:

An equivalent problem is $$\max_{(a,b,c)\ne 0} \frac{(3a+28b+35c)(20a+23b+33c)}{a^2 + b^2 + c^2}$$ or, equivalently $$\max_{a^2 + b^2 + c^2 =1} (3a+28b+35c)(20a+23b+33c)$$

The symmetric quadratic form $$(a,b,c) \mapsto (3a+28b+35c)(20a+23b+33c)=\\=60 a^2 + 629 a b + 799 a c + 644 b^2 + 1729 b c + 1155 c^2$$ is given by the symmetric matrix

$$\left( \begin{matrix} 60 & 629/2& 799/2 \\629/2 & 644 & 1729/2 \\ 799/2 &1729/2 &1155 \end{matrix}\right)$$

with eigenvalues $\frac{3877}{2}$, $-\frac{159}{2}$ and $0$. By the theory, the above maximum is $\frac{3877}{2}=1938.5$. So the best estimate for the problem is $a^2+b^2+c^2 \ge \frac{1}{1938.5}$