how to prove the following inequality by using real analysis

Question

how to prove this inequality by real anlysis

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

I finally came with the following inequality. but I dont know what to do after that

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

Answer 1

Note that the function $f(t) = \frac{t}{1+t}$ is increasing for $t\geq 0$.

So, using

• $|x+y| \leq |x| + |y|$ and
• $\frac{|x|}{1+|x|} \geq \frac{|x|}{1+|x|+|y|}$ and $\frac{|y|}{1+|y|} \geq \frac{|y|}{1+|x|+|y|}$

you get

$$\frac{|x|}{1+|x|} + \frac{|y|}{1+|y|} \geq \frac{|x| + |y|}{1+|x| +|y|} \stackrel{f\; increasing}{\geq}\frac{|x+y|}{1+|x+y|}$$

Answer 2

Let $f(t):= \frac{t}{1+t}$ for $t \ge 0.$ Show that $f$ is increasing. Since $|x+y| \le |x|+|y|$, we get

$\frac{|x+y|}{1+|x+y|}=f(|x+y|) \le f(|x|+|y|)=\frac{|x|+|y|}{1+|x|+|y|}$.

Can you proceed ?

Answer 3

Just use the triangle inequality.

$$ \frac{|x+y|}{1+|x|+|y|} \leq \frac{|x|+|y|}{1+|x|+|y|}=\frac{|x|}{1+|x|+|y|}+\frac{|y|}{1+|x|+|y|}\leq \frac{|x|}{1+|x|}+\frac{|y|}{1+|y|} $$