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I would like to have an explanation of the following solution to the following problem (instead of another solution), and a similar problem has been asked by me before.

Let $a_1$, $a_2$, $a_3$, $a_4$, $a_5$ be real numbers such that $$a_1a_2+a_2a_3+a_3a_4+a_4a_5+a_5a_1=20,\\a_1a_3+a_2a_4+a_3a_5+a_4a_1+a_5a_2=24.$$ Prove that $a_1^2+a_2^2+a_3^2+a_4^2+a_5^2\ge22+2\sqrt5$.

The solution defined $z_i=\frac1{\sqrt5}\sum\limits_{k=1}^5a_k\omega^{-ki}$, but it did not state what $\omega$ was, perhaps $\mathrm e^{\frac{2\pi}3\mathrm i}$.

Then it claimed that \begin{align} \sum\limits_{i=1}^5a_i^2&=\sum\limits_{i=1}^5|z_i|^2,\\\sum\limits_{i=1}^5a_ia_{i+1}&=\sum\limits_{i=1}^5\cos\frac{2k\pi}5|z_i|^2,\tag1\\\sum\limits_{i=1}^5a_ia_{i+2}&=\sum\limits_{i=1}^5\cos\frac{4k\pi}5|z_i|^2.\tag2 \end{align} I don't understand why any of the equations hold.

Then it finished the proof by considering $2\cos\frac{2\pi}5(1)-2\cos\frac\pi5(2)+\sum\limits_{i=1}^5|z_i|^2$.

Can anyone explain how this solution works?

youthdoo
  • 3,663

1 Answers1

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Choose $\omega=e^{2\pi i/5}$, we verify the three equations.

First equation: \begin{align*} \sum_{j=1}^5|z_j|^2&=\frac15\sum_{j=1}^5\left|\sum_{k=1}^5a_k\omega^{-jk}\right|^2\\ &=\frac15\sum_{j=1}^5\sum_{1\leq k,l\leq5}a_ka_l\omega^{j(l-k)}\\ &=\sum_{k=1}^5a_k^2+\frac15\sum_{1\leq k\neq l\leq5}a_ka_l\underbrace{\left(\omega^{l-k}+\omega^{2(l-k)}+\omega^{3(l-k)}+\omega^{4(l-k)}+\omega^{5(l-k)}\right)}_{=\,0}. \end{align*}

Second equation: \begin{align*} &\sum_{j=1}^5\cos\left(\frac{2j\pi}5\right)|z_j|^2=\text{Re}\sum_{j=1}^5\omega^j|z_j|^2\\ &=\frac15\text{Re}\sum_{j=1}^5\sum_{1\leq k,l\leq5}a_ka_l\omega^{j(l-k-1)}\\ &=\sum_{k=1}^5a_ka_{k+1}+\frac15\text{Re}\!\!\!\!\!\sum_{\substack{1\leq k,l\leq5\\l-k\not\equiv1\ \text{mod}\ 5}}\!\!\!\!\!a_ka_l\underbrace{\left(\omega^{l-k-1}+\omega^{2(l-k-1)}+\omega^{3(l-k-1)}+\omega^{4(l-k-1)}+\omega^{5(l-k-1)}\right)}_{=\,0}. \end{align*}

Third equation is similar (replacing $\omega^j$ with $\omega^{2j}$).

Now with $(1),(2)$ denoting the second and third equation respectively, we have

\begin{align*} &2\cos\frac{2\pi}5(1)-2\cos\frac\pi5(2)+\sum\limits_{j=1}^5|z_j|^2\\ =\ &\sum_{j=1}^5\underbrace{\left(2\cos\frac{2\pi}5\cos\frac{2j\pi}5-2\cos\frac\pi5\cos\frac{4j\pi}5+1\right)}_{=\ A_j}|z_j|^2. \end{align*}

Using trigonometry identities, we get $$A_j=\frac{\sqrt5-1}2\cos\frac{2j\pi}5-\frac{\sqrt5+1}2\cos\frac{4j\pi}5+1$$ where $j=1,2,...,5$. It is easy to show $A_j$ is minimised when $j=2$, with $A_2=0$. Hence $A_j\geq0$ and

$$\sum_{j=1}^5a_j^2=\sum_{j=1}^5|z_j|^2\geq2\cos\frac\pi5(2)-2\cos\frac{2\pi}5(1)=22+2\sqrt5.$$

Mengchun Zhang
  • 2,995