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solakis1
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prove the following inequality:
$\dfrac{|a+b|}{1+|a+b|}$ $\leq \dfrac{|a|}{1+|a|}$ +$\dfrac{|b|}{1+|b|}$
$\dfrac{|a+b|}{1+|a+b|}$ $\leq \dfrac{|a|}{1+|a|}$ +$\dfrac{|b|}{1+|b|}$
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]we have;solakis said:prove the following inequality:
$\dfrac{|a+b|}{1+|a+b|}$ $\leq \dfrac{|a|}{1+|a|}$ +$\dfrac{|b|}{1+|b|}$
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]
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.
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.
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.
One common mistake made when using the proof of inequality $|a+b| \leq |a| + |b|$ is assuming that the inequality holds true for all values of a and b. This is not always the case, as there are certain conditions and restrictions that need to be considered. Another mistake is forgetting to take into account the sign of the numbers, as the absolute value of a negative number is equal to its positive value.
The proof of inequality $|a+b| \leq |a| + |b|$ can be applied in various areas of science, such as physics, chemistry, and economics. In physics, it is used to represent the uncertainty principle in quantum mechanics. In chemistry, it is used to determine the strength of a chemical bond. In economics, it is used to analyze the relationship between supply and demand.