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

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And for the 2nd inequality from the top instead of $|a+b|\leq |a|+|b|$ should be:$|a+b|\leq |a|+|b|+|a||b|$I apologize for the mistake.
  • #1
solakis1
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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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  • #2
to ifdahl[sp] ifdahl in the same way you proved the post : interesting inequality you can prove the above inequality[/sp]
 
  • #3
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:

\[\frac{\left | a+b \right |}{1+\left | a+b \right |}\leq \frac{\left | b \right |}{1+\left | b \right |}\leq \frac{\left | a \right |}{1+\left | a \right |}+\frac{\left | b \right |}{1+\left | b \right |}\].
 
  • #4
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:
$$\frac{|a+b|}{1+|a+b|} \le \frac{|a|}{1+|a+b|} + \frac{|b|}{1+|a+b|} \le \frac{|a|}{1+|a|} + \frac{|b|}{1+|b|}$$
 
  • #5
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]
 
  • #6
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]
 
  • #7
[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]
 
  • #8
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]
 
  • #9
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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What is the definition of "Proof of Inequality: $|a+b| \leq |a| + |b|$"?

The proof of inequality $|a+b| \leq |a| + |b|$ is a mathematical statement that shows the relationship between the absolute values of two numbers, a and b. It states that the absolute value of the sum of two numbers is always less than or equal to the sum of the absolute values of those numbers.

Why is "Proof of Inequality: $|a+b| \leq |a| + |b|$" important in mathematics?

The proof of inequality $|a+b| \leq |a| + |b|$ is important in mathematics because it is a fundamental concept that is used in various mathematical proofs and applications. It helps in understanding the properties of absolute values and inequalities, and is also used in solving equations and inequalities involving absolute values.

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One real-life example of the proof of inequality $|a+b| \leq |a| + |b|$ is the distance between two points on a coordinate plane. The absolute value of the difference between the x-coordinates and y-coordinates of two points is always less than or equal to the distance between those two points. Another example is the triangle inequality, where the sum of the lengths of any two sides of a triangle is always greater than or equal to the length of the third side.

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