Proof fraction with sum of numerators and denominators is larger than the minimum of the individual fraction

Question

For my research, I am dealing with the following hypothesis:

$\frac{x_1+x_2}{d_1+d_2} \geq \min \left(\frac{x_1}{d_1}, \frac{x_2}{d_2}\right)$

with the following assumptions

• $x_1,x_2,d_1,d_2 \in \mathbb{N}$ with $d_1+d_2 \geq 1$
• $0 \leq \frac{x_1+x_2}{d_1+d_2} \leq 1$

As for the proof, I did not get any further than below:

Let's assume that $\min \left(\frac{x_1}{d_1}, \frac{x_2}{d_2}\right) = \frac{x_1}{d_1}$, then

$\frac{x_1+x_2}{d_1+d_2} \geq \min \left(\frac{x_1}{d_1}, \frac{x_2}{d_2}\right) = \frac{x_1}{d_1}$

Multiplying both sides with the denominator gives us

$d_1(x_1+x_2) \geq x_1(d_1+d_2)$

Dividing both sides with the shorter terms gives us

$\frac{x_1+x_2}{x_1} \geq \frac{d_1+d_2}{d_1}$

Any suggestions for how I could proceed are much appreciated!

Answer 1

So assume $\frac {x_1}{d_1}\le \frac {x_2}{d_2}$. As wall are positive that is true if and only if $x_1d_2 \le x_2d_1$.

So how does $\frac {x_1+x_2}{d_1+d_2}$ compare to $\frac {x_i}{d_i}$.

Let's use $???$ to mean $\le$ or $\ge$ but we don't know which but every step is an if and only iff. In other words, lets "solve for ????". Or maybe I should call this a "work backwords" proof.

(WARNING!!!!! "work backwards" proofs only work if every step is "if and only if" or "only if". The fail if any step is an actual "if ... then" step. Because we are working backward our steps need to all be "we can only get then .... if....".)

$\frac {x_1+x_2}{d_1+d_2} ??? \frac {x_i}{d_i} \iff$

$x_1d_i + x_2d_i ??? x_id_1 + x_id_2$

If $i = 1$ we get

$x_1d_1 + x_2d_1 ???? x_1d1 + x_1d_2$

$x_2d_1 ??? x_1d_2$.

As $x_2d_1 \ge x_1d_2$ we have $\frac {x_1+x_2}{d_1+d_2} \ge \frac {x_1}{d_1}=\min(\frac {x_i}{d_i})$.

ANd if $i=2$ we get by the exact same argument

$x_1d_2 + x_2d_2 ???? x_2d1 + x_2d_2$

$x_1d_2 ??? x_2d_1$.

As $x_1d_2 \le x_2d_1$ we have $\frac {x_1+x_2}{d_1+d_2} \le \frac {x_2}{d_2}=\max(\frac {x_i}{d_i})$.

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If you don't like that we can work forward (now that we can see where the secret step came from:

$\frac {x_1}{d_1}\le \frac {x_2}{d_2}$

$x_1d_2 \le x_2d_1$

$x_1d_1 + x_1d_2 \le x_1d_1 + x_2d_1$ (this was the secret step)

$x_1(d_1 + d_2) \le d_1(x_1 + x_2)$

$\frac {x_1}{d_1} \le \frac {x_1+x_2}{d_1+d_2}$.

Answer 2

Most of the stated assumptions can be dropped; $x_1, x_2, d_1, d_2$ can be any real numbers with $d_1 > 0,$ $d_2 > 0.$

Let $$ \alpha = \min\left\{\frac{x_1}{d_1}, \frac{x_2}{d_2}\right\}. $$ Then $$ \frac{x_1}{d_1} \geqslant \alpha, \ \frac{x_2}{d_2} \geqslant \alpha. $$ But $d_1 > 0$ and $d_2 > 0,$ therefore $$ x_1 \geqslant \alpha d_1, \ x_2 \geqslant \alpha d_2. $$ Therefore $$ x_1 + x_2 \geqslant \alpha d_1 + \alpha d_2 = \alpha(d_1 + d_2). $$ But $d_1 + d_2 > 0,$ therefore $$ \frac{x_1 + x_2}{d_1 + d_2} \geqslant \alpha = \min\left\{\frac{x_1}{d_1}, \frac{x_2}{d_2}\right\}. \quad \square $$

Answer 3

We have for positive real numbers:

\begin{align} &\frac{a}{b}\le \frac{a+c}{b+d}\\ \iff\\ &a(b+d) \le b(a+c)\\ \iff\\ &a b+a d \le ba + bc\\ \iff\\ &ad \le b c\\ \iff\\ &\frac{a}{b} \le \frac{c}{d} \end{align}