# Use Triangle Inequalities to Prove an Inequality [duplicate]

Use the triangle inequalities to prove that:

$$\frac{|a+b|}{1+|a+b|}\le\frac{|a|}{1+|a|} + \frac{|b|}{1+|b|}$$ for all $a,b\in\mathbb{R}$.

I've made it this far, but it doesn't seem to be helping much; what I end up with at the end isn't very useful:

Clearly I'm taking a wrong turn somewhere, but I'm not sure what else to try. Any advice?

## marked as duplicate by Michael Rozenberg calculus StackExchange.ready(function() { if (StackExchange.options.isMobile) return; $('.dupe-hammer-message-hover:not(.hover-bound)').each(function() { var$hover = $(this).addClass('hover-bound'),$msg = $hover.siblings('.dupe-hammer-message');$hover.hover( function() { $hover.showInfoMessage('', { messageElement:$msg.clone().show(), transient: false, position: { my: 'bottom left', at: 'top center', offsetTop: -7 }, dismissable: false, relativeToBody: true }); }, function() { StackExchange.helpers.removeMessages(); } ); }); }); Feb 13 '18 at 5:50

• You aren't considering that $|a|+|b| \ge |a|$. So you are closer than you think. – fleablood Feb 13 '18 at 5:43
$$\frac{|a+b|}{1+|a+b|}=1-\frac{1}{1+|a+b|}\leq1-\frac{1}{1+|a|+|b|}=$$ $$=\frac{|a|}{1+|a|+|b|}+\frac{|b|}{1+|a|+|b|}\leq\frac{|a|}{1+|a|}+\frac{|b|}{1+|b|}.$$