Induction of inequality involving AP

Prove by induction that $$(a_{1}+a_{2}+\cdots+a_{n})\left(\frac{1}{a_{1}}+\frac{1}{a_{2}}+\cdots+\frac{1}{a_{n}}\right)\geq n^{2}$$

where $n$ is a positive integer and $a_1, a_2,\dots, a_n$ are real positive numbers

Hence, show that $$\csc^{2}\theta +\sec^{2}\theta +\cot^{2}\theta \geq 9\cos^{2}\theta$$

Please help me.

Thank you!

Define $x = a_1 + a_2 \cdots + a_{n-1}$ and $y = \frac{1}{a_1} + \frac{1}{a_2} \cdots + \frac{1}{a_{n-1}}$.

By the induction hypothesis, we have $xy \geq (n-1)^2$.

We need to prove $(x + a_n)(y + \frac1{a_n}) \geq n^2$

But clearly by induction hypothesis, $(x + a_n)(y + \frac1{a_n}) = xy + ya_{n} + \frac{x}{a_n} + 1 \geq (n-1)^2 + ya_{n} + \frac{x}{a_n} + 1$

So it suffices to prove: $ya_n + \frac{x}{a_n} \geq 2n - 2 $

Now use the definition of $x$ and $y$, and use $\displaystyle \frac{a_i}{a_n} + \frac{a_n}{a_i} \geq 2$ to obtain $ya_n + \frac{x}{a_n} \geq 2n - 2 $

For the first part, we need a base case and an inductive step.

Base Case: suppose we have only one number, $a_1$. Then $$(a_1)\left(\frac1{a_1}\right)=1\leq 1$$

Inductive Step: By Isomorphism's work, this amounts to showing that $$ a_n\left(\frac{1}{a_1} + \frac{1}{a_2} \cdots + \frac{1}{a_{n-1}}\right) + \frac1{a_n}(a_1 + a_2 \cdots + a_{n-1})\geq 2n-2 $$ Start by multiplying through to get $$ \begin{align} a_n\left(\frac{1}{a_1} + \frac{1}{a_2} \cdots + \frac{1}{a_{n-1}}\right) + \frac1{a_n}(a_1 + a_2 \cdots + a_{n-1}) & = \\ \left(\frac{a_n}{a_1} + \frac{a_n}{a_2} \cdots + \frac{a_n}{a_{n-1}}\right) + \left(\frac{a_1}{a_n} + \frac{a_2}{a_n} \cdots + \frac{a_{n-1}}{a_n}\right) &= \\ \left(\frac{a_n}{a_1} + \frac{a_1}{a_n}\right) + \left(\frac{a_n}{a_2} + \frac{a_2}{a_n}\right) + \dots + \left(\frac{a_n}{a_{n-1}} + \frac{a_{n-1}}{a_n}\right) &\geq \\ 2+2+\dots+2 \end{align} $$

If $a,u,v>0$, then $$\begin{align}(u+a)(v+\frac1a)&=uv+av+\frac ua+1\\&=(uv+2\sqrt{uv}+1)+(av-2\sqrt{uv}+\frac ua)\\&=(\sqrt{uv}+1)^2+(\sqrt {av}-\sqrt{\frac ua})^2\\&\ge (\sqrt{uv}+1)^2\end{align}$$ Therfore we can proceed by induction: If $(a_1+\ldots+a_n)(\frac1{a_1}+\ldots \frac1{a_n})\ge n^2$, let $u=a_1+\ldots+a_n$, $v=\frac1{a_1}+\ldots \frac1{a_n}$ and $a=a_{n+1}$ in the above to find $(a_1+\ldots+a_n+a_{n+1})(\frac1{a_1}+\ldots \frac1{a_n}+\frac1{a_{n+1}})\ge (n+1)^2$ as desired. Together with the trivial base case $n=1$ (i.e. $a_1\cdot \frac1{a_1}\ge 1^2$), the claim follows for all $n\ge 1$.

If "prove by induction" is a requirement of the problem, it is a meaningless requirement, because any proof of $T(n)$ a fortiori proves "if $T(k)$ holds for all $k\lt n$ then $T(n)$". So let's forget about induction.

With the usual convention about adding indices modulo $n$, since $(a_1+a_2+\cdots+a_n)\left(\frac1{a_1}+\frac1{a_2}+\cdots+\frac1{a_n}\right)=\sum_{k=1}^n\left(\frac{a_1}{a_{1+k}}+\frac{a_2}{a_{2+k}}+\cdots+\frac{a_n}{a_{n+k}}\right)$, it will be enough to show that $\frac{a_1}{a_{1+k}}+\frac{a_2}{a_{2+k}}+\cdots+\frac{a_n}{a_{n+k}}\ge n$. But this follows immediately from the arithmetic-geometric mean inequality: since the geometric mean of those $n$ terms is clearly equal to $1$, their arithmetic mean is at least $1$, i.e., their sum is at least $n$.