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solakis1
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Prove for all $n\in N$
$\dfrac{|a_1+...a_n|}{1+|a_1+...+a_n|}\leq\dfrac{|a_1|}{1+|a_1|}+...\dfrac{|a_n|}{1+|a_n|}$
$\dfrac{|a_1+...a_n|}{1+|a_1+...+a_n|}\leq\dfrac{|a_1|}{1+|a_1|}+...\dfrac{|a_n|}{1+|a_n|}$
[sp]To follow your way We have:Opalg said:First step is to prove the inequality when $n=2$ and both the numbers are positive. So we want to show that if $a,b\geqslant0$ then $$\frac{a+b}{1+a+b}\leqslant \frac a{1+a} + \frac b{1+b}.$$ Write that as $$1 - \frac1{1+a+b} \leqslant 1-\frac1{1+a} + 1 - \frac1{1+b},$$ $$ \frac1{1+a} + \frac1{1+b} \leqslant 1 + \frac1{1+a+b} = \frac{2+a+b}{1+a+b}.$$ Multiply out the fractions to get $$(2+a+b)(1+a+b) \leqslant (2+a+b)(1+a+b+ab).$$ That last inequality is clearly true, and all the steps are reversible. Therefore the first inequality is true.
The next step is to prove the given inequality in the case when $a_1,\ldots,a_n$ are all positive. This is done by induction, the base case $n=2$ being covered by Step 1 above. For the inductive step, suppose that $$\frac{a_1 + \ldots + a_{n-1}}{1+a_1 + \ldots + a_{n-1}} \leqslant \frac{a_1}{1+a_1} + \ldots + \frac{a_{n-1}}{1+a_{n-1}},$$ and apply the Step 1 inequality with $a=a_1+\ldots+a_{n-1}$ and $b=a_n$ to get $$\frac{a_1 + \ldots + a_n}{1+a_1 + \ldots + a_n} \leqslant \frac{a_1 + \ldots + a_{n-1}}{1+a_1 + \ldots + a_{n-1}} + \frac{a_n}{1+a_n}.$$ The result then follows from the inductive hypothesis.
Finally, suppose that $a_1,\ldots,a_n$ are arbitrary real (or even complex) numbers. For $x\geqslant0$, the function $\dfrac x{1+x}$ is an increasing function. Since $|a_1+\ldots+a_n| \leqslant |a_1| + \ldots + |a_n|$ it follows that $$\frac{|a_1 + \ldots + a_n|}{1+|a_1 + \ldots + a_n|} \leqslant \frac{|a_1| + \ldots + |a_n|}{1+|a_1| + \ldots + |a_n|}.$$ It then follows from Step 2 applied to the positive numbers $|a_1|,\ldots,|a_n|$ that $$\frac{|a_1 + \ldots + a_n|}{1+|a_1 + \ldots + a_n|} \leqslant \frac{|a_1| + \ldots + |a_n|}{1+|a_1| + \ldots + |a_n|} \leqslant \frac{|a_1|}{1+|a_1|} + \ldots + \frac{|a_n|}{1+|a_n|}.$$
The proof of the Triangle Inequality for n Natural Numbers is a mathematical proof that shows that the sum of any two sides of a triangle is always greater than the third side. It is a fundamental property of triangles and is used in various fields of mathematics and science.
The Triangle Inequality is important because it helps us determine whether a set of numbers can form a triangle or not. It is also used in various geometric and algebraic proofs, and it has many applications in fields such as physics, engineering, and computer science.
The Triangle Inequality for n Natural Numbers is proved using mathematical induction. This method involves proving the statement for a base case (usually n=3) and then showing that if the statement holds for n=k, it also holds for n=k+1. By proving these two steps, we can conclude that the statement holds for all natural numbers.
The Triangle Inequality has many real-life applications, such as determining the shortest distance between two points, optimizing travel routes, and designing structures that can withstand certain forces. It is also used in the study of inequalities and optimization problems in mathematics and economics.
Yes, there are variations of the Triangle Inequality for n Natural Numbers, such as the Reverse Triangle Inequality, which states that the absolute value of the difference between two sides of a triangle is always less than or equal to the sum of the other two sides. There are also generalizations of the Triangle Inequality for other types of polygons and higher dimensions.