This question is from Bak & Newman's question 15, page 160:
$\mathbf{Q:}$ Show that $|(z-1)^2(z+1)|\leq \frac{16\sqrt{3}}{9}$ whenever $|z|=1$. Then deduce that
$$|\sum\limits_{k=0}^n (-1)^k \binom{n}{k}\binom{2n}{k}|\leq (\frac{16\sqrt 3}{9})^n $$
I tried the first part, but am clueless as to how I can get the $\sqrt 3$ out..
For the second part, I arrived at the step:
$$ |(z-1)^2(z+1)|^n =|(1+z)(1-z)^2|^n =|(1+z)^n(1-z)^{2n}|=|(\sum\limits_{k=0}^n \binom{n}{k}z^k)(\sum\limits_{k=0}^{2n}\binom{2n}{k}(-1)^kz^k)|$$
So my question for this part is how can I manipulate the $2n$ in the second sum such that I am able to "combine" the sums to get the desired inequality?
About the first part: by letting $z=e^{i\theta}$ we have $$ \|(z-1)^2(z+1)\|^2 = (2+2\cos\theta)(2-2\cos\theta)^2 \tag{1}$$ and the derivative of the RHS vanishes iff $\sin\theta=0$ or $\cos\theta\in\left\{-\frac{1}{3},1\right\}$. By straightforward computations the maximum modulus of $(z-1)^2(z+1)$ over the unit circle is $\frac{16}{9}\sqrt{3}$.
About the second part: $$ \sum_{k=0}^{n}\binom{n}{n-k}\binom{2n}{k}(-1)^k = [x^n]\left(\sum_{j=0}^{n}\binom{n}{j}x^j\right)\cdot\left(\sum_{j=0}^{2n}\binom{2n}{j}(-1)^j x^j\right) \tag{2}$$ hence the LHS is the coefficient of $x^n$ in the product between $(1+x)^n$ and $(1-x)^{2n}$, or the coefficient of $x^n$ in $\left((1-x)^2 (1+x)\right)^n$. By Cauchy's integral formula such coefficient is given by $$ \frac{1}{2\pi i}\oint_{\|z\|=1}\frac{\left((1-z)^2(1+z)\right)^{n}}{z^{n+1}}\,dz \tag{3}$$ and the modulus of $(3)$, by the triangle inequality and the first part, is bounded by $\left(\frac{16}{9}\sqrt{3}\right)^n$.