, 2020

Solution:
We first prove the following lemma:
Lemma. For positive reals $a, b, c, d$, the inequality
$$\frac{a^{3 / 2}}{c^{1 / 2}} + \frac{b^{3 / 2}}{d^{1 / 2}} \geq \frac{(a+b)^{3 / 2}}{(c+d)^{1 / 2}}$$
holds.
Proof. Apply Hölder's inequality in the form
$$\left(\frac{a^{3 / 2}}{c^{1 / 2}} + \frac{b^{3 / 2}}{d^{1 / 2}}\right)^{2}(c+d) \geq (a+b)^{3}.$$
For $k \geq 2$, applying the lemma to $a = (k-1)^{2}$, $b = 8k + 8$, $c = y_{k-1}^{2}$, $d = y_{k}^{2} - y_{k-1}^{2}$ yields
$$\frac{(k-1)^{3}}{y_{k-1}} + \frac{(8k+8)^{3 / 2}}{\sqrt{y_{k}^{2} - y_{k-1}^{2}}} \geq \frac{(k+3)^{3}}{y_{k}}.$$
We also have the equality
$$\frac{(8 \cdot 1 + 8)^{3 / 2}}{\sqrt{y_{1}^{2} - y_{0}^{2}}} = \frac{(1+3)^{3}}{y_{1}}.$$
Summing the inequality from $k=2$ to $k=n$ with the equality yields
$$\sum_{k=1}^{n} \frac{(8k+8)^{3 / 2}}{\sqrt{y_{k}^{2} - y_{k-1}^{2}}} \geq \sum_{k=1}^{n} \frac{(k+3)^{3} - k^{3}}{y_{k}} + \frac{n^{3}}{y_{n}} \geq \sum_{k=1}^{n} \frac{9\left(k^{2} + 3k + 3\right)}{y_{k}}$$
Hence the inequality holds for $\alpha = \frac{16 \sqrt{2}}{9}$. In the reverse direction, this is sharp when $y_{n} = n(n+1)(n+2)(n+3)$ (so that $y_{k-1} = \frac{k-1}{k+3} y_{k}$ for $k=2, \ldots, n$) and $n \rightarrow \infty$.