# inequality

#### solakis

##### Active member
prove the following inequality:

$\dfrac{|a+b|}{1+|a+b|}$ $\leq \dfrac{|a|}{1+|a|}$ +$\dfrac{|b|}{1+|b|}$

#### solakis

##### Active member
to ifdahl

ifdahl in the same way you proved the post : interesting inequality you can prove the above inequality

#### lfdahl

##### Well-known member
Case I: $a$ and $b$ have the same sign:

Let the function $f$ be defined by: $f(x) = \frac{\left | x \right |}{1+\left | x \right |}$. Obviously $f$ is even, and $f’(x)$ is not defined in $x=0$, but $f$ is differentiable in the two domains $\mathbb{R}_-$ and $\mathbb{R}_+$, and we have by inspection: $f’’(x) < 0$ in both domains. Thus $f$ is concave on both sides of the ordinate.

Jensens inequality with equal weights then gives us:

$f\left ( \frac{a+b}{2} \right ) \leq \frac{1}{2}\left ( f(a) + f(b)\right ) \\ \frac{\frac{1}{2}\left | a+b \right |}{1+\frac{1}{2}\left | a+b \right |} \leq \frac{1}{2}\left ( \frac{\left | a \right |}{1+\left | a \right |} +\frac{\left | b \right |}{1+\left | b \right |} \right )$ - or

$\frac{\left | a+b \right |}{1+\frac{1}{2}\left | a+b \right |} \leq \frac{\left | a \right |}{1+\left | a \right |} +\frac{\left | b \right |}{1+\left | b \right |}$, which immediately implies:

$\frac{\left | a+b \right |}{1+\left | a+b \right |} \leq \frac{\left | a \right |}{1+\left | a \right |} +\frac{\left | b \right |}{1+\left | b \right |}$.

Case II: $a$ and $b$ have opposite sign. Here, we cannot use the concavity argument, but the inequality is still valid:

First note, that: $\left | a+b \right |\leq max\left \{ \left | a \right |,\left | b \right | \right \}$

WLOG let $\left | b \right | \geq \left | a+b \right |$. Denote $x= \left | a+b \right |, \Delta = \left | b \right |-x \geq 0$:

The inequality: $\frac{\left | a+b \right |}{1+\left | a+b \right |}=\frac{x}{1+x} \leq \frac{x+\Delta }{1+x+\Delta } = \frac{\left | b \right |}{1+\left | b \right |}$ is true because:

$\frac{x+\Delta }{1+x+\Delta }-\frac{x}{1+x}= \frac{(1+x)(x+\Delta )-x(1+x+\Delta )}{(1+x)(1+x+\Delta )}=\frac{\Delta }{(1+x)(1+x+\Delta )}\geq 0$. Thus the inequality holds, from which we immediately have:

$\frac{\left | a+b \right |}{1+\left | a+b \right |}\leq \frac{\left | b \right |}{1+\left | b \right |}\leq \frac{\left | a \right |}{1+\left | a \right |}+\frac{\left | b \right |}{1+\left | b \right |}$.

#### Klaas van Aarsen

##### MHB Seeker
Staff member
The triangle inequality tells us that $|a+b|\le |a|+|b|$. So:
$$\frac{|a+b|}{1+|a+b|} \le \frac{|a|}{1+|a+b|} + \frac{|b|}{1+|a+b|}$$
Furthermore, positive fractions are smaller if their denominator is bigger. It means that $\frac{1}{1+|a+b|}\le \frac 1{1+|a|}$ and also $\frac{1}{1+|a+b|}\le \frac 1{1+|b|}$.
Therefore:
$$\frac{|a+b|}{1+|a+b|} \le \frac{|a|}{1+|a+b|} + \frac{|b|}{1+|a+b|} \le \frac{|a|}{1+|a|} + \frac{|b|}{1+|b|}$$

#### solakis

##### Active member
The triangle inequality tells us that $|a+b|\le |a|+|b|$. So:
$$\frac{|a+b|}{1+|a+b|} \le \frac{|a|}{1+|a+b|} + \frac{|b|}{1+|a+b|}$$
Furthermore, positive fractions are smaller if their denominator is bigger. It means that $\frac{1}{1+|a+b|}\le \frac 1{1+|a|}$ and also $\frac{1}{1+|a+b|}\le \frac 1{1+|b|}$.
Therefore:
$$\frac{|a+b|}{1+|a+b|} \le \frac{|a|}{1+|a+b|} + \frac{|b|}{1+|a+b|} \le \frac{|a|}{1+|a|} + \frac{|b|}{1+|b|}$$
for a=3 and for b=-:3 does it not the inequality : $\frac{1}{1+|a+b|}\le \frac 1{1+|a|}$ become : $1\leq \dfrac{1}{4}$

#### solakis

##### Active member
prove the following inequality:

$\dfrac{|a+b|}{1+|a+b|}$ $\leq \dfrac{|a|}{1+|a|}$ +$\dfrac{|b|}{1+|b|}$
we have;

$\dfrac{|a+b|}{1+|a+b|}$ $\leq\dfrac{|a|(1+|b|)+|b|(1+|a|)}{(1+|a|)(1+|b|)}$
or
$|a+b|(1+|a|)(1+|b|)\leq(1+|a+b|)[|a|(1+|b|)+|b|(1+|a|)]$
or
$|a+b|(1+|a|+|b|+|a||b|)\leq (1+|a+b|)(|a|+|b|+2|a||b|)$
or $|a+b|+|a||a+b|+|b||a+b|+|a||b||a+b|\leq |a|+|b|+2|a||b|+|a||a+b|+|b||a+b|+2|a||b||a+b|$
And after canceling terms we end up with:
$|a+b|\leq |a|+|b|+2|a||b|+|a||b||a+b|$
Wich is true because :
$|a+b|\leq |a|+|b|$
$0\leq2|a||b| =|ab|$
$0\leq |a||b||a+b|=|ab(a+b)|$
Bebause $|X|\geq 0$for all real X

#### solakis

##### Active member
$|a+b|\leq |a|+|b|$
$0\leq 2|a||b|$
$0\leq 2|a||b|$

$|a+b|\leq |a|+|b|+2|a||b|+|a||b||a+b|$

Now adding to both sides$|a||a+b|$.................$|b||a+b|$...................................$|a||b||a+b|$
we have:

$|a+b|+|a||a+b|+|b||a+b|+|a||b||a+b|\leq |a|+|b|+2|a||b|+|a||a+b|+|b||a+b|+2|a||b||a+b|$

Taking common factor $|a+b|$ in both sides we have:

$|a+b|(1+|a|+|b|+|a||b|)\leq |a|+|b|+2|a||b|+|a+b|( |a|+|b|+2|a||b|)$............................................................(1)

Now $1+|a|+|b|+|a||b|=1+|a|+|b|(1+|a|)=(1+|a|)(1+|b|)$...................................................................................(2)

And substituting (2) into (1) we have:

$|a+b|(1+|a|)(1+|b|)\leq |a|+|b|+2|a||b|+|a+b|( |a|+|b|+2|a||b|)$..........................................................................(3)

And taking common factor:$|a|+|b|+2|a||b|$ in the left side (3) becomes:

$|a+b|(1+|a|)(1+|b|)\leq(|a|+|b|+2|a||b|)(1+|a+b|)$..............................................................................(4)

But $1+|a+b|>0\Rightarrow\dfrac{1}{1+|a+b|}>0$.............................................................................(5)

So we can multiply (4) by (5) and we have:

$\dfrac{|a+b|(1+|a|)(1+|b|)}{1+|a+b|}\leq |a|+|b|+2|a||b|$....................................................(6)

But $|a|+|b|+2|a||b|=|a|+|b|+|a||b|+|a||b|=|a|(1+|b|)+|b|(1+|a|)$.........................................................(7)

And substituting (6) into (7) into (6) we have:

$\dfrac{|a+b|(1+|a|)(1+|b|)}{1+|a+b|}\leq |a|(1+|b|)+|b|(1+|a|)$.....................................................................(8)

But $1+|a|>0$ also $1+|b|>0$ hence $(1+|a|)(1+|b|)>0\Rightarrow\dfrac{1}{(1+|a|)(1+|b|)}>0$....................................(9)

So we can multiply (8) by (9) and the result is:

$\dfrac{|a+b|}{1+|a+b|}\leq\dfrac{|a|(1+|b|)+|b|(1+|a|)}{(1+|a|)(1+|b|)}$

$=\dfrac{|a|}{1+|a|} +\dfrac{|b|}{1+|b|}$[/sp

#### solakis

##### Active member
$|a+b|\leq |a|+|b|$
$0\leq 2|a||b|$
$0\leq 2|a||b||a+b|$

$|a+b|\leq |a|+|b|+2|a||b|+|a||b||a+b|$

Now adding to both sides$|a||a+b|$.................$|b||a+b|$...................................$|a||b||a+b|$
we have:

$|a+b|+|a||a+b|+|b||a+b|+|a||b||a+b|\leq |a|+|b|+2|a||b|+|a||a+b|+|b||a+b|+2|a||b||a+b|$

Taking common factor $|a+b|$ in both sides we have:

$|a+b|(1+|a|+|b|+|a||b|)\leq |a|+|b|+2|a||b|+|a+b|( |a|+|b|+2|a||b|)$............................................................(1)

Now $1+|a|+|b|+|a||b|=1+|a|+|b|(1+|a|)=(1+|a|)(1+|b|)$...................................................................................(2)

And substituting (2) into (1) we have:

$|a+b|(1+|a|)(1+|b|)\leq |a|+|b|+2|a||b|+|a+b|( |a|+|b|+2|a||b|)$..........................................................................(3)

And taking common factor:$|a|+|b|+2|a||b|$ in the left side (3) becomes:

$|a+b|(1+|a|)(1+|b|)\leq(|a|+|b|+2|a||b|)(1+|a+b|)$..............................................................................(4)

But $1+|a+b|>0\Rightarrow\dfrac{1}{1+|a+b|}>0$.............................................................................(5)

So we can multiply (4) by (5) and we have:

$\dfrac{|a+b|(1+|a|)(1+|b|)}{1+|a+b|}\leq |a|+|b|+2|a||b|$....................................................(6)

But $|a|+|b|+2|a||b|=|a|+|b|+|a||b|+|a||b|=|a|(1+|b|)+|b|(1+|a|)$.........................................................(7)

And substituting (6) into (7) into (6) we have:

$\dfrac{|a+b|(1+|a|)(1+|b|)}{1+|a+b|}\leq |a|(1+|b|)+|b|(1+|a|)$.....................................................................(8)

But $1+|a|>0$ also $1+|b|>0$ hence $(1+|a|)(1+|b|)>0\Rightarrow\dfrac{1}{(1+|a|)(1+|b|)}>0$....................................(9)

So we can multiply (8) by (9) and the result is:

$\dfrac{|a+b|}{1+|a+b|}\leq\dfrac{|a|(1+|b|)+|b|(1+|a|)}{(1+|a|)(1+|b|)}$

$=\dfrac{|a|}{1+|a|} +\dfrac{|b|}{1+|b|}$

#### solakis

##### Active member
Sorry there is a terrible typo in both of my two prεvious solutions;
For the 3rd inequality from the top instead of $0\leq 2|a||b|$ or $0\leq 2|a||b||a+b|$ it shoulb be:

$0\leq |a||b||a+b|$