Inequalities

Show that $|x|+|y| \leq |x+y|+|x-y| \quad and \quad \frac{|x+y|}{1+|x+y|} \leq \frac{|x|}{1+|x|} + \frac{|y|}{1+|y|} \quad for \quad x,y \in \mathbb{R}$ We have $\quad|x+y+x-y| \leq |x+y|+|x-y|$ $\implies 2|x| \leq |x+y|+|x-y| \dots A$ Now, $|x+y+y-x| \leq |x+y|+|y-x|$ $\implies 2|y| \leq |x+y|+|-1||x-y|$ $\implies 2|y| \leq |x+y|+|x-y| \dots B$ Adding $A$ and $B$ we get $|x|+|y| \leq |x+y|+|x-y|$ The second inequality holds $iff$ $\frac{(1+|x|)(1+|y|)|x+y|}{1+|x+y|} \leq |x|(1+|y|)+|y|(1+|x|)$ $iff \quad (1+|x|)(1+|y|)|x+y| \leq (|x|+|y|+2|xy|)(1+|x+y|)$ (multiply and cancel out terms) $iff \quad |x+y| \leq |x|+|y| +|xy|(2+|x+y|)$ Now for any real $x$ and $y$, $|xy|(2+|x+y|) \geq 0$ and we know that $|x+y| \leq |x|+|y|$ Thus the last inequality holds, which in turn proves the original ineqality.