0

To prove : if $a>0,b>0,x>0,a<b $, then $\dfrac{a}{b}<\dfrac{a+x}{b+x}<1$

Here's How I go about solving such things.

Before starting, I'll split and clarify. Therefore, Restating it:

$\dfrac{a}{b}<\dfrac{a}{b+x}+\dfrac{x}{b+x}<1$

Firstly I'll try to prove that $\dfrac{a}{b}<\dfrac{a}{b+x}+\dfrac{x}{b+x}$

I would rephrase the statement as "is $\dfrac{a}{b+x}$ sufficient to add to $\dfrac{x}{b+x}$ to make the above statement true?"

For that, We'll find out what the condition of the additive part has to be such that the above statement holds:

$\dfrac{a}{b}<d+\dfrac{x}{b+x}$

$\dfrac{a}{b}-\dfrac{x}{b+x}<d$

$\dfrac{a(b+x)-bx}{b(b+x)}<d$

Now if I can prove that the part I'm adding $\dfrac{a}{b+x}$, is indeed greater than $\dfrac{a(b+x)-bx}{b(b+x)}$, that would suffice.

Now, if it were true that $\dfrac{a(b+x)-bx}{b(b+x)}<\dfrac{a}{b+x}$

Then it would also follow (as $b>0$) that $\dfrac{a(b+x)-bx}{b(b+x)}<\dfrac{ab}{b(b+x)}$

Also would follow, $a(b+x)-bx<ab$

So if I can prove that $a(b+x)-bx<ab$, then with inverse opertions, the original statement would be automatically proved.

$a(b+x)-bx<ab \Leftrightarrow ab+ax-bx<ab \Leftrightarrow ab+x(a-b)<ab$

as $a<b$, So, $x(a-b)$ is always negative with $x>0$, hence proved.

2nd part

$\dfrac{a}{b+x}+\dfrac{x}{b+x}<1$

As, $b,x>0$

$\dfrac{a}{b+x}<1-\dfrac{x}{b+x} \Leftrightarrow \dfrac{a}{b+x}<\dfrac{b}{b+x} \Leftrightarrow a<b$

As given $a<b$, Hence proved as well.

Please give feedback on way of thinking, and the proof.

math_adventure
  • 21
  • 5
    Extremely complicated way of doing things. Just cross-multiply and see what the inequalities really say. You will finish in two lines. – Kavi Rama Murthy May 06 '22 at 09:38
  • You're right but I happened to have approached it in a quite elementary manner, cross multiply didn't came in mind at that time. – math_adventure May 06 '22 at 09:49

1 Answers1

6

While correct, there's a much more straightforward way to prove that statement:

First, since $x>0$ and $0<a<b$ we have $$ 0<a+x<b+x $$ therefore $$ \frac{a+x}{b+x}<1 $$

Second, since $x>0$ and $0<a<b$ we have $$ a(b+x)=ab+ax<ab+bx=b(a+x) $$ therefore (since $b>0$ and $b+x>0$) $$ \frac{a}{b}<\frac{a+x}{b+x} $$

Mor A.
  • 1,006