Proving an interesting property of absolute values: $\frac{|x+y|}{1+|x+y|}\le\frac{|x|}{1+|y|}+\frac{|y|}{1+|y|}$ [duplicate]

Question

If $x$ and $y$ are real numbers, show that

$$\frac{|x+y|}{1+|x+y|}≤\frac{|x|}{1+|y|}+\frac{|y|}{1+|y|}$$

I am having difficulty in proving this equation. I don't know where to start. Your help will be highly appreciated.

Solution 1:

We know that $|x+y|\leq|x|+|y|$. You need to prove $$\frac{|x+y|}{1+|x+y|}\leq\frac{|x|}{1+|x|}+\frac{|y|}{1+|y|}$$ $$or,~|x+y|.(1+|x|)(1+|y|)\leq(1+|x+y|)(|x|.(1+|y|)+|y|.(1+|x|))$$ by cross multiplication. By expanding we get $$|x+y|\leq|x|+|y|+2|x||y|+|x||y||x+y|$$ which is obviously true since $2|x||y|+|x||y||x+y|\geq0$.

Solution 2:

Consider the function $$ f(t)=\frac{t}{1+t} $$ where $t \ge 0$. It is immediate to see that it is an increasing function on its domain. Now, since $|x+y| \le |x|+|y|$, you will have $$ \frac{|x+y|}{1+|x+y|}\le \frac{|x|}{1+|x|+|y|}+\frac{|y|}{1+|x|+|y|} $$ and, since $\frac{1}{1+|x|+|y|} \le \frac{1}{1+|x|}$, you can conclude.

P.S. Observe that the above argument holds also for the inequality you write... in the comment, I write the classical inequality students must solve in their analysis courses!