<s>[INST] Show $\frac{n^2}{2^{\sqrt{\log(n)}}} \geq \frac{n}{2}. $ So we know that $\frac{n^2}{2} \geq \frac{n}{2}$, but I'm stuck proving that $\frac{n^2}{2} \geq \frac{n^2}{2^{\sqrt{\log n}}}\geq \frac{n}{2}$. Am I missing something?
 [/INST] Let $\sqrt {\log n}=u,\thinspace u≥0$, then we have
$$\begin{align}&\frac{n^2}{2^{\sqrt{\log(n)}}} ≥ \frac{n}{2}\\
\iff &\frac{n^2}{2^{\sqrt{\log(n)}}} ≥ \frac{n^2}{2n}\\
\iff &n≥2^{\sqrt{\log(n)}-1}\\
\iff &e^{u^2}≥2^{u-1}\\
\iff &e^{u^2}>e^{u-1}>2^{u-1}.\end{align}$$
This means, the equality is impossible.
Therefore, we have
$$\frac{n^2}{2^{\sqrt{\log(n)}}} > \frac{n}{2}.$$
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