MHB Proof of Inequality: $|a+b| \leq |a| + |b|$

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The discussion centers on proving the inequality $\dfrac{|a+b|}{1+|a+b|} \leq \dfrac{|a|}{1+|a|} + \dfrac{|b|}{1+|b|}$. The proof involves manipulating the terms and applying properties of absolute values to demonstrate that the left-hand side is less than or equal to the right-hand side. Key steps include establishing that $|a+b| \leq |a| + |b|$ and utilizing the positivity of certain terms to maintain the inequality. The conclusion confirms that the inequality holds true under the given conditions. Overall, the proof effectively illustrates the relationship between the absolute values of sums and individual terms.
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prove the following inequality:
$\dfrac{|a+b|}{1+|a+b|}$ $\leq \dfrac{|a|}{1+|a|}$ +$\dfrac{|b|}{1+|b|}$
 
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to ifdahl[sp] ifdahl in the same way you proved the post : interesting inequality you can prove the above inequality[/sp]
 
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:

\[\frac{\left | a+b \right |}{1+\left | a+b \right |}\leq \frac{\left | b \right |}{1+\left | b \right |}\leq \frac{\left | a \right |}{1+\left | a \right |}+\frac{\left | b \right |}{1+\left | b \right |}\].
 
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:
$$\frac{|a+b|}{1+|a+b|} \le \frac{|a|}{1+|a+b|} + \frac{|b|}{1+|a+b|} \le \frac{|a|}{1+|a|} + \frac{|b|}{1+|b|}$$
 
Klaas van Aarsen said:
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:
$$\frac{|a+b|}{1+|a+b|} \le \frac{|a|}{1+|a+b|} + \frac{|b|}{1+|a+b|} \le \frac{|a|}{1+|a|} + \frac{|b|}{1+|b|}$$

[sp]for a=3 and for b=-:3 does it not the inequality : $\frac{1}{1+|a+b|}\le \frac 1{1+|a|}$ become : $1\leq \dfrac{1}{4}$[/sp]
 
solakis said:
prove the following inequality:
$\dfrac{|a+b|}{1+|a+b|}$ $\leq \dfrac{|a|}{1+|a|}$ +$\dfrac{|b|}{1+|b|}$
[sp]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[/sp]
 
[sp]Let's start with:
$|a+b|\leq |a|+|b|$
$0\leq 2|a||b|$
$0\leq 2|a||b|$
adding the above we have:

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

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[/sp]
 
solakis said:
[sp]Let's start with:
$|a+b|\leq |a|+|b|$
$0\leq 2|a||b|$
$0\leq 2|a||b||a+b|$
adding the above we have:

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

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]
 
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:

$0\leq |a||b||a+b|$
 
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