Proof that $\sum a_k a_{k+1} \leq \dfrac{a^2}{4} $ where $a = \sum a_k $

Question

Let $a_k > 0$ ( $k=1,...,n$ ) nad set $a = a_1 + ... +a_n $. Prove that

$$ \sum_{k=1}^{n-1} a_k a_{k+1} \leq \dfrac{a^2}{4} $$

Proof.

Notice that the LHS is equivalent to $\sum_{k=1}^{n-1} a_{k+1}(a_{k}+a_{k+2}) $ and now one can apply AM-GM inequality to each two positive terms to obtain

$$ \sum_{k=1}^{n-1} a_{k+1}(a_{k}+a_{k+2}) \leq \dfrac{ a_2^2 + ... + a_{n-1}^2 + (a_1+a_3)^2 + ... + (a_{n-2}+a_n)^2}{2} = \dfrac{ \sum_{k=1}^n a_k^2 + 2\sum_{k=1}^{n-2} a_ka_{k+2}}{2}$$

Is there a way to bound this expression by $a^2/4$? OR am I on the wrong track altogether?

Answer 1

Use the following AM-GM: $$\sum_{k=1}^{n-1}a_ka_{k+1}\leq(a_1+a_3+...)(a_2+a_4+..)\leq\left(\frac{\sum\limits_{k=1}^na_k}{2}\right)^2=\frac{a^2}{4}.$$