Induction absolute values proof verification

Question

Prove that $$\frac{|a_1+a_2+\cdots+a_n|}{1+|a_1+a_2+\cdots+a_n|}\leq\frac{|a_1|}{1+|a_1|}+\frac{|a_2|}{1+|a_2|}+\cdots+\frac{|a_n|}{1+|a_n|}$$ Here is my proof I think that my solution is too simple. Denote:$$a_1+a_2+\cdots+a_n=x$$ For basis $n=1$ it's true.
This is induction hypothesis: $$\frac{|x|}{1+|x|}+\frac{|a_{n+1}|}{1+|a_{n+1}|}\geq\frac{|x+a_{n+1}|}{1+|x+a_{n+1}|}$$ We rewrite it as: $$\frac{|x|(1+|a_{n+1}|)+|a_{n+1}|(1+|x|)}{(1+|x|)(1+|a_{n+1}|)}\geq\frac{|x+a_{n+1}|}{1+|x+a_{n+1}|}$$ And then cross multiply it $$|x|+2|x||a_{n+1}|+|a_{n+1}|+|x||x+a_{n+1}|+2|x||a_{n+1}||x+a_{n+1}|+|a_{n+1}||x+a_{n+1}|\geq|x+a_{n+1}|+|a_{n+1}||x+a_{n+1}|+|x||x+a_{n+1}|+|x||a_{n+1}||x+a_{n+1}|$$ After we reduce the same terms it is $$|x|+2|x||a_{n+1}|+|a_{n+1}|+|x||a_{n+1}||x+a_{n+1}|\geq|x+a_{n+1}|$$ Since $$|x|+|a_{n+1}|\geq|x+a_{n+1}|$$ We strengthen the hypothesis $$2|x||a_{n+1}|+|x||a_{n+1}||x+a_{n+1}|+|x+a_{n+1}|\geq|x+a_{n+1}|$$ $$1+\frac{2|a_{n+1}||x|}{|x+a_{n+1}|}+|x||a_{n+1}|\geq1 $$$$ \frac{2|a_{n+1}||x|}{|x+a_{n+1}|}+|x||a_{n+1}|\geq0$$ And that's true because we have all absolute values on the LHS.

Answer 1

Your proof is not a proof by inducition.

Say you have a statement about integers, $P(n)$, and you want to prove this statement.

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Structure of a proof by induction:

1. You prove that $P(1)$ is a true statement
2. You prove that $P(n)\implies P(n+1)$ is a true statement.

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Structure of your proof:

1. You prove (it's pretty obvious) that $P(1)$ is true
2. You prove that $P(n+1)$ is true.

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So, your proof isn't really a proof by induction, because it doesn't follow the correct structure.

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That said, your proof is also wrong because

$$\frac{|x|}{1+|x|}+\frac{|a_{n+1}|}{1+|a_{n+1}|}\geq\frac{|x+a_{n+1}|}{1+|x+a_{n+1}|}$$

is not what the induction hypothesis is. The induction hypothesis is

$$\frac{|a_1+a_2+\cdots+a_n|}{1+|a_1+a_2+\cdots+a_n|}\leq\frac{|a_1|}{1+|a_1|}+\frac{|a_2|}{1+|a_2|}+\cdots+\frac{|a_n|}{1+|a_n|}$$

and the statement you want to prove is

$$\frac{|a_1+a_2+\cdots+a_n+a_{n+1}}{1+|a_1+a_2+\cdots+a_n+a_{n+1}|}\leq\frac{|a_1|}{1+|a_1|}+\frac{|a_2|}{1+|a_2|}+\cdots+\frac{|a_n|}{1+|a_n|}+\frac{|a_{n+1}|}{1+|a_{n+1}|}$$

Which translates into

$$\frac{|a_1|}{1+|a_1|}+\frac{|a_2|}{1+|a_2|}+\cdots+\frac{|a_n|}{1+|a_n|}+\frac{|a_{n+1}|}{1+|a_{n+1}|}\geq \frac{|x+a_{n+1}|}{1+|x+a_{n+1}|}$$

Now, it may be that you are simply missing a step, but that in itself is a mistake.

Answer 2

I have a simple proof.

Proof: Without lose of generality we can assume $a_{i}\neq 0$ for $i=1,\cdots,n$

by Cauchy-Schwarz inequality and notice the function $f(x)=\frac{x}{1+x}$ is increase for $x\geq 0$,we have \begin{align*} \sum_{i=1}^{n}\frac{|a_{i}|}{1+|a_{i}|}&=\sum_{i=1}^{n}\frac{|a_{i}|^{2}}{|a_{i}|^2+|a_{i}|}\\ &\geq \frac{\left(\sum_{i=1}^{n}|a_{i}|\right)^2}{\sum_{i=1}^{n}|a_{i}|+\sum_{i=1}^{n}|a_{i}|^{2}}\\ &\geq \frac{\left(\sum_{i=1}^{n}|a_{i}|\right)^2}{\sum_{i=1}^{n}|a_{i}|+\left(\sum_{i=1}^{n}|a_{i}| \right)^{2}}\\ &=\frac{\sum_{i=1}^{n}|a_{i}|}{1+\sum_{i=1}^{n}|a_{i}|}\\ &\geq \frac{|a_{1}+a_{2}+\cdots+a_{n}|}{1+|a_{1}+a_{2}+\cdots+a_{n}|} \end{align*} Hence we are done!

Answer 3

@5xum

Here is a graphical proof of the property:

$$\tag{1} \forall a_1,a_2,\cdots a_n \in \mathbb{R}, \ f(\sum_k a_k)\leq \sum_k f(a_k)$$

with $f(x):=\dfrac{|x|}{1+|x|}$ (graphical representation in red on the figure below).

This is NOT a rigorous proof ; it is only intended to give a strong intuitive flavor, close to evidence, to property (1).

Let us first mention some remarkable properties $f$:

• (i) $f(0)=0 \ ;$

• (ii) $\forall x>0, f(x)>0 \ ;$

• (iii) $f$ is an even function ;

• (iv) $f$ is derivable (but in 0) with decreasing $|f'(x)|$ from $1$ in $0_+$ to $0$ at $\infty.$

Consider the figure below. Let us call the red curve (the curve of $f$) a "flower". Let $A_0(0,0), \ A_1(a_1,f(a_1))$ and more generally $A_n(\sum_k a_k,\sum_k f(a_k)).$

Property (1) can be graphically interpreted in the following way: in each new point $A_k$ let a new flower blossom... (i.e., by a translation $\vec{A_0A_k}$, construct a copy of the initial curve). Due to the properties of $f$ mentionned ipwars, it is impossible that any new flower meets its support flower ; as a consequence, by an immediate recursion, (1) holds.

It remains to transform this graphical argument (that I wish convincing) into a formal proof...

Remark: As a consequence of this reasoning, other functions could be taken in place of function $f$ as defined above. It suffices that such a function verifies properties (i) to (iv) (one could even lower the conditions on (iv)).