Nice induction exercise

I prove it assuming $a_i\ge 0$ (this can be rectified, see below). Interpret the given expression as a quadratic in $a_n$: \[ a_n^2 - (2a_{n-1})a_n + a_{n-1}^2 - 2a_{n-1}a_{n-2}\ge 0, \]whose discriminant and roots are \[ \Delta = 8a_{n-1}a_{n-2}\quad\text{and}\quad r_{1,2} = a_{n-1} \pm \sqrt{2a_{n-1}a_{n-2}}. \]From here, we find, using the facts $a_n$ is increasing and $\sqrt{x}\ge 0$ for all $x$, that \[ a_n\ge a_{n-1} + \sqrt{2a_{n-1}a_{n-2}},\quad\forall n\ge 3. \]We now set the stage for strong induction. The case $n=3$ is trivial, so assume $a_{n-1}\ge S_{n-2}$ and $a_{n-2}\ge S_{n-3}$ for $S_k=a_1+\cdots+a+k$. Notice that \begin{align*} a_n\ge a_{n-1} + \sqrt{2a_{n-2}S_{n-2}} = a_{n-1}+\sqrt{2a_{n-2}^2 + 2a_{n-2}S_{n-3}} \ge a_{n-1}+\sqrt{2a_{n-2}^2 + 2S_{n-3}^2}, \end{align*}where the first inequality uses $a_{n-1}\ge S_{n-2}$ and the second uses $a_{n-2}\ge S_{n-3}$. Now, using $2u^2+2v^2\ge (u+v)^2$ valid through Cauchy, we get that \[ \sqrt{2a_{n-2}^2 + 2S_{n-3}^2}\ge a_{n-2}+S_{n-3}. \]Combining last two displays, we arrive at \[ a_n\ge a_{n-1}+a_{n-2}+S_{n-3} = S_{n-1}, \]as claimed. Assumption $a_i\ge 0$. Note that if $a_i<0$ for all $i$, we have a clear contradiction: since $a_i$ is increasing, $a_i-a_{i-1}\ge 0$ so if $a_i<0$ for all $i$, we immediately obtain that the given condition cannot be satisfied. Now, let $i$ be the smallest index with $a_i\ge 0$. Clearly $a_k\ge 0$ for all $k\ge i$ and moreover $i=3$: if $i\ge 4$, then using the given condition for $i-1$ we again have a contradiction. Hence, $a_k\ge 0$ for all $k\ge 3$. One can simply start the sequence from $k=3$, drop $a_1,a_2$ etc. I skip this.

Weirdish inequality. We have two cases. The first being the case where $(a_n) < 0$ for all $n\in \mathbb{N}$. In this case the problem is trivial since for all $n\in \mathbb{N}$, \[a_n > a_{n-1} \geq a_{n-1} + a_{n-2}+ \dots + a_1\]the first inequality due to $(a_n)$ being strictly increasing and the second due to $(a_n)$ being negative. We are then left to deal with the case where there exists some $n\in \mathbb{N}$ such that $a_n >0$. In this case, say $a_m$ is the smallest positive term of the given sequence $(a_n)$. Clearly, for all $n\leq m+1$, \[a_n > a_n-1 \geq a_{n-1} + a_{n-2} + \dots + a_1 \]similar to the all negatives case. Further, for all $n\geq m+2$, we see that we have the inequality \[a_n(a_n-2a_{n-1})+a_{n-1}(a_{n-1}-2a_{n-2}) \geq 0 \text{ where } a_n , a_{n-1} , a_{n-2} \in \mathbb{R}^+ \]Upon rearrangement this gives us that \begin{align*} a_n^2 - 2a_na_{n-1} + a_{n-1}^2 & \geq 2a_{n-1}a_{n-2}\\ (a_n - a_{n-1})^2 & \geq 2a_{n-1}a_{n-2}\\ a_n - a_{n-1} & \geq \sqrt{2a_{n-1}a_{n-2}} \end{align*}So, we wish to show \[\sqrt{2a_{n-1}a_{n-2}} \geq a_{n-2}+ \dots + a_1\]which is trivial if $a_{n-2} + \dots a_1<0$. Else, we proceed via induction. Note that, \begin{align*} \sqrt{2a_{n-2}a_{n-1}} & \geq \sqrt{(a_{n-2}+a_{n-3}+\dots + a_1)(a_{n-2}+\dots + a_1)} \\ &= a_{n-2}+\dots + a_1 \end{align*}by applying the strong inductive hypothesis on $a_{n-1}$ and $a_{n-2}$. Thus, the result is true for $a_n$ as well, which completes the induction.