Indian Statistical Institute B.Math & B.Stat Solved Problems, Vinod Singh ~ Kolkata
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} $
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We have $\quad|x+y+x-y| \leq |x+y|+|x-y| $
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$ \implies 2|x| \leq |x+y|+|x-y| \dots A$
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Now, $|x+y+y-x| \leq |x+y|+|y-x|$
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$ \implies 2|y| \leq |x+y|+|-1||x-y| $
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$\implies 2|y| \leq |x+y|+|x-y| \dots B$
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Adding $A$ and $B$ we get $ |x|+|y| \leq |x+y|+|x-y|$
The second inequality holds $$$$
$iff$
$\frac{(1+|x|)(1+|y|)|x+y|}{1+|x+y|} \leq |x|(1+|y|)+|y|(1+|x|)$
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$ iff \quad (1+|x|)(1+|y|)|x+y| \leq (|x|+|y|+2|xy|)(1+|x+y|)$
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(multiply and cancel out terms)
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$iff \quad |x+y| \leq |x|+|y| +|xy|(2+|x+y|)$
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Now for any real $x$ and $y$, $|xy|(2+|x+y|) \geq 0$ and we know that $|x+y| \leq |x|+|y|$
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Thus the last inequality holds, which in turn proves the original ineqality.
