HMMT February

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 $a=(k-1)^{2}, b=8 k+8, c=y_{k-1}^{2}, d=y_{k}^{2}-y_{k-1}^{2}$ yields $\frac{(k-1)^{3}}{y_{k-1}}+\frac{(8 k+8)^{3 / 2}}{\sqrt{y_{k}^{2}-y_{k-1}^{2}}} \geq \frac{(k+3)^{3}}{y_{k}}$. 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 $\sum_{k=1}^{n} \frac{(8 k+8)^{3 / 2}}{\sqrt{y_{k}^{2}-y_{k-1}^{2}}} \geq \sum_{k=1}^{n} \frac{9\left(k^{2}+3 k+3\right)}{y_{k}}$. 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 $n \rightarrow \infty$.