<s>[INST] Prove by induction the following inequality for all n∈N $\frac1{\sqrt{1}} + \frac1{\sqrt{2}}+\frac1{\sqrt{3}}+...+\frac1{\sqrt{x}}\ge {\sqrt{x}}$
I proved the basic case: and realize it is equal to 1, but I have absolutely no idea how to create prove the left and right side using the induction hypothesis.
Please help :) I've solved equality's before using induction, but this is the first time I've done an inequality, and although I'm sure the process is fairly similar, I think the square root is messing me up.
 [/INST] Let $s_n$ represent the sum up to $n$, and suppose that $s_n \ge \sqrt{n}$ for induction. We want to show that $s_{n + 1} \ge \sqrt{n + 1}$; to this end, note
\begin{align*}
s_{n + 1} - s_n &= \frac{1}{\sqrt{n + 1}}
\end{align*}
Therefore,
\begin{align*}
s_{n + 1} &= \frac{1}{\sqrt{n + 1}} + s_n \\
&\ge \frac{1}{\sqrt{n + 1}} + \sqrt{n} \\
&= \frac{1 + \sqrt{n}\sqrt{n + 1}}{\sqrt{n + 1}} \\
&\ge \frac{1 + \sqrt{n} \sqrt{n}}{\sqrt{n + 1}} \\
&= \frac{n + 1}{\sqrt{n + 1}} = \sqrt{n + 1}
\end{align*}
as desired. The first inequality is by the induction hypothesis, and the second is just by noting that
$$\sqrt{n + 1} \ge \sqrt{n}$$
</s>