How to prove that $\frac{a^2 + b^2 -2}{a + b - 2} < 1 + b$ if $a < b$ and $a, b > 1$?

Question

How to prove that $\frac{a^2 + b^2 -2}{a + b - 2} < 1 + b$ if $a < b$ and $a, b > 1$? I tried some numbers and it seems that is it true, but how to prove that for all posible $\{a, b\}$?

Answer 1

Let $a=1+x$ and $b=1+y$.

Thus, $$1+b-\frac{a^2+b^2-2}{a+b-2}=2+y-\frac{x^2+y^2+2x+2y}{x+y}=\frac{x(y-x)}{x+y}>0.$$

Answer 2

If we get rid of fraction we get $$a^2-a<ab-b$$ and this is true since $$a(a-1)<b(a-1)$$ is equivalent to $a<b$

Answer 3

your inequality is equivalent to $$a^2+b^2-2<(1+b)(a+b-2)$$ and this is equivalent to $$(1-a)(a-b)>0$$ this is true!

Answer 4

We have $$\begin{align}\frac{a^2 + b^2 -2}{a + b - 2} < 1 + b&\Leftarrow a^2+b^2-2<(1+b)(a+b-2)=a-b-2+ab+b^2\\&\Leftarrow a^2<a-b+ab\Leftarrow a(a-1)<b(a-1)\Leftarrow a<b\end{align}$$

Note that $a+b-2>0$ and we can divide by $a-1$.

Answer 5

Oh, just muck about.

$a,b> 1$ so $a + b -2 > 0$ so

$\frac x{a+b-2} < y \iff x < y(a+b-2)$.

So is $a^2 + b^2 - 2 < (1+b)(a+b-2)$?

Well, $(1+b)(a+b -2) = a+b -2 + ab + b^2 - 2b$ which.... geez...

Well $a^2 + b^2 - 2< a+b - 2 + ab + b^2 - 2b \iff a^2 < a-b +ab$

And .... well, $b > a$ so $a^2 < ab$ but we also have $a-b < 0$ so I don't think we have it. unless $ab - a^2 > b-a$....

But $ab- a^2 = a(b-a)$ and as $a > 1$ and $b-a > 0$, $a(b-a) > b-a$

.... and that does it!

To put it all together:

$\frac {a^2 + b^2 - 2}{a + b - 2} < 1+ b \iff $

$a^2 + b^2 - 2 < (1+b)(a+b - 2)= a- b + ab -2$ (note: this is only true because $a + b - 2> 0$.

$\iff a^2 < a-b + ab \iff$

$ab -a^2 = a(b-a) > b-a $ (and as $b > a$ and $b-a > 0$)

$\iff a > 1$ (which was a given.)

So, yes, it is true.

Big things can come from just mucking about.

...

Heck, we can even look like geniuses pulling stuff out of thin air:

$a > 1$ and $b > a$ so $b-a > 0$

and $a(b-a) > b-a$

So $ab - a^2 > b-a$

$a^2 < a-b + ab$

$a^2 + b^2 - 2 < a + b - 2 + ab + b^2 - 2b = (1+b)(a + b -2)$

As $a,b > 1$ we know $a+ b - 2 > 0$. So

$\frac {a^2 + b^2 -2}{a+b -2} < 1+ b$.

Wow, that looked like magic!

====

Then there is brute force:

$\frac {a^2 + b^2 - 2}{a+b-2} = 1 + \frac {a^2 + b^2 - 2-a-b+2}{a+b-2}=1 + \frac {a^2 + b^2 -a-b}{a+b-2}=$

$1 + b + \frac {a^2 + b^2 -a-b -ab - b^2 +2b}{a+b-2}=1+ b + \frac {a^2 -a -ab +b}{a+b-2}=$

$1+b + \frac {a(a-1) -b(a-1)}{a+b-2} = 1+b + \frac {(a-b)(a-1)}{a+b-2}$

And $a-1 >0$ and $a-b < 0$ and $a+b -2 > 1+1 -2 = 0$.

So $\frac {(a-b)(a-1)}{a+b-2} < 0$.

So $1+b + \frac {(a-b)(a-1)}{a+b-2} < 1+b$.