Proof of Triangle Inequality for $n$ Natural Numbers

[solakis1]

Prove for all $n\in N$

$\dfrac{|a_1+...a_n|}{1+|a_1+...+a_n|}\leq\dfrac{|a_1|}{1+|a_1|}+...\dfrac{|a_n|}{1+|a_n|}$

 

[Opalg]

Spoiler

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 $$\frac{|a_1 + \ldots + a_n|}{1+|a_1 + \ldots + a_n|} \leqslant \frac{|a_1| + \ldots + |a_n|}{1+|a_1| + \ldots + |a_n|} \leqslant \frac{|a_1|}{1+|a_1|} + \ldots + \frac{|a_n|}{1+|a_n|}.$$

 

[solakis1]

Spoiler

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 $$\frac{|a_1 + \ldots + a_n|}{1+|a_1 + \ldots + a_n|} \leqslant \frac{|a_1| + \ldots + |a_n|}{1+|a_1| + \ldots + |a_n|} \leqslant \frac{|a_1|}{1+|a_1|} + \ldots + \frac{|a_n|}{1+|a_n|}.$$

[sp]To follow your way We have:

$(a+b)\leq a+b$ Divide both sides by 1+a+b>0 and we have:

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

Now for the induction.

Since $(a_1+.....a_n)\leq a_1+.......a_n)$

Divide both sides by $(1+a_1.......a_n)$ and you have the desired result

Now substitute all the a's with there corresponding absolute value since they are positive and we have the desired inequality which holds for all values of a's[/sp]