Proof by induction for positive real numbers

Question

I am trying to, for self-study, prove the following fact for positive real numbers.

Let $a_1,\dots,a_n,b_1,\dots,b_n \in \mathbb{R}_{>0}$ be positive real numbers. Then, one of the following two inequalities holds: $$ \frac{a_1}{b_1} + \cdots + \frac{a_n}{b_n} \geq n \hspace{10pt} \mathrm{or} \hspace{10pt} \frac{b_1}{a_1} + \cdots + \frac{b_n}{a_n} \geq n. $$

Here is my attempt at a solution.

Let's proceed by induction on $n$. Let $n = 1$. Then we either have $a_1 \geq b_1$ or $a_1 \leq b_1$. If $a_1 \geq b_1$, then $\frac{a_1}{b_1} \geq 1$. If $a_1 \leq b_1$, then $\frac{b_1}{a_1} \geq 1$.

Suppose the inequality holds for $n$. We will show it for $n + 1$. Let $a_1, \ldots, a_{n}, a_{n+1}, b_1, \ldots, b_n, b_{n+1} \in \mathbb{R}_{>0}$ be positive. By the induction hypothesis, we either have $$ \sum\limits_{i=1}^n \frac{a_i}{b_i} \geq n \hspace{10pt} \mathrm{or} \hspace{10pt} \sum\limits_{i=1}^n \frac{b_i}{a_i} \geq n. $$ Suppose first that $\sum\limits_{i=1}^n \frac{a_i}{b_i} \geq n$. If $a_{n+1} \geq b_{n+1}$, then $\frac{a_{n+1}}{b_{n+1}} \geq 1$, and so $$ \sum\limits_{i=1}^{n+1} \frac{a_i}{b_i} \geq n + 1. $$ If $a_i \leq b_i$, then $0 < \frac{a_i}{b_i} \leq 1$, so. $$ \sum\limits_{i=1}^{n+1} \frac{a_i}{b_i} = \sum\limits_{i=1}^{n} \frac{a_i}{b_i} + \frac{a_{n+1}}{b_{n+1}} > n, $$ so $\sum\limits_{i=1}^{n+1} \frac{a_i}{b_i} \geq n+1$. By swapping labels, the exact same argument shows that if $\sum\limits_{i=1}^n \frac{b_i}{a_i} \geq n$, then $\sum\limits_{i=1}^{n+1} \frac{b_i}{a_i} \geq n + 1$.

How does this proof look?

Answer 1

Your proof is currently not correct. In the case where $$\sum_{i=1}^n\frac{a_i}{b_i}\ge n\quad\text{and}\quad\frac{a_{n+1}}{b_{n+1}}<1$$ you say that then $$\sum_{i=1}^{n+1}\frac{a_i}{b_i}\ge n+1,$$ but this is not necessarily true. To give a concrete example (and build intuition), you can consider $a_1=3,b_1=2,a_2=1,b_2=2$.

Answer 2

For $x>-1$ (so dividing by $1+x$ do not swap inequality) then consider this result:

$$1-x^2\le 1\implies 1-x\le \frac 1{1+x}$$

From there let's define $\dfrac{a_i}{b_i}=1+u_i\quad$ and $\quad U=\sum\limits_{i=1}^n u_i$

Note that $a_i,b_i>0\implies u_i>-1$

We have $\begin{cases} \displaystyle\sum\limits_{i=1}^n \frac{a_i}{b_i}=\sum\limits_{i=1}^n(1+u_i)=n+U\\ \displaystyle\sum\limits_{i=1}^n \frac{b_i}{a_i}=\sum\limits_{i=1}^n \frac 1{1+u_i}\ge \sum\limits_{i=1}^n(1-u_i) = n-U \end{cases}$

Therefore it is obvious that depending on the sign of $U$, one of these two expressions is greater than $n$.

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Here is an alternate proof, for $x>0$ we have $(x-1)^2\ge 0\implies x+\frac 1x\ge 2$

Therefore $\displaystyle\sum\limits_{i=1}^n \frac{a_i}{b_i}+\sum\limits_{i=1}^n \frac{b_i}{a_i}=\sum\limits_{i=1}^n \Big(\frac{a_i}{b_i}+\frac{b_i}{a_i}\Big)\ge\sum\limits_{i=1}^n2=2n$

So one of the two sums is greater than $n$, indeed if $S_1+S_2\ge 2n$ then

• either $S_1\ge n$
• or $S_1\le n\implies -S_1\ge -n\implies S_2\ge 2n-S_1\ge 2n-n\ge n$