Define the length of a polynomial $A = a_0 + a_1x+\cdots +a_nx^n \in \mathbb{R}[x]$ to be $$L(A) = \sum_{i=0}^n |a_i|$$, and let $$\lambda (r,s) : = \inf \frac{L(AB)}{L(A)L(B)}$$ where the degree of $A$ is at most $r$, and the degree of $B$ is at most $s$.
Prove that $\lambda(r,r) \leq \frac{1}{2^r}$.
I've tried two ideas, neither of which I could really get to work. The first was an induction argument, in which I was able to successfully show the base case, but couldn't get the inductive step.
The second thought I had was to see if I could just explicitly find polynomials $A$ and $B$ that have length $(1/2)^r$ for any $r$, that would also work to show the infimum is less then that value.
Any hints or thoughts would be greatly appreciated.
For a fixed $r\in\mathbb{Z}_{\geq0}$, let $A(x):=(1+x)^r$ and $B(x):=(1-x)^r$. Then, $$L(A)=L(B)=L(AB)=2^r\,.$$ That is, $$\frac{L(AB)}{L(A)\,L(B)}=\frac{1}{2^r}\,.$$ Hence, $$\lambda(r,r)\leq \frac{1}{2^r}\,.$$
There are some easy-to-prove results. First, $\lambda(r,0)=\lambda(0,r)=1$ for all $r\in\mathbb{Z}_{\geq 0}$. Next, $$\lambda(r,1)=\lambda(1,r)\leq \dfrac{1}{r+1}$$ for all $r\in\mathbb{Z}_{>0}$, where the equality holds, at least, for $r=1$. Finally, for all positive integers $r$ and $s$, we get $$\lambda(r,rs)=\lambda(rs,r)\leq \dfrac{1}{(s+1)^r}\,.$$