+0  
 
0
150
2
avatar+16 

I tried by starting with (x+y)^2+(x-y)^2>=(x-y)^2, which brought me to 2x^2+2y^2>= (|x|-|y|)^2 , have no idea what else to try. Thanks!

AspiringActuary  Jul 26, 2017
edited by Guest  Jul 26, 2017

Best Answer 

 #1
avatar+18777 
+3

Using triangle inequity prove |x+y|>=||x|-|y||

 

\(\begin{array}{|rcll|} \hline |x+y| & \stackrel{?} \ge & ||x|-|y|| \\ \hline \end{array}\)

 

\(\begin{array}{|lrcll|} \hline &|y| = |(x+y)-x| & \le & |x+y| + |x| \\ & |y| & \le & |x+y| + |x| \\ (1) & |y|-|x| & \le & |x+y| \\\\ & |x| = |(x+y)-y| & \le & |x+y| + |y| \\ & |x| & \le & |x+y| + |y| \\ (2) & |x|-|y| & \le & |x+y| \\\\ \hline (1) \text{ and } (2) & ||x|-|y| | & \le & |x+y| \\ & |x+y| & \ge & ||x|-|y| | \\ \hline \end{array}\)

 

laugh

heureka  Jul 26, 2017
Sort: 

2+0 Answers

 #1
avatar+18777 
+3
Best Answer

Using triangle inequity prove |x+y|>=||x|-|y||

 

\(\begin{array}{|rcll|} \hline |x+y| & \stackrel{?} \ge & ||x|-|y|| \\ \hline \end{array}\)

 

\(\begin{array}{|lrcll|} \hline &|y| = |(x+y)-x| & \le & |x+y| + |x| \\ & |y| & \le & |x+y| + |x| \\ (1) & |y|-|x| & \le & |x+y| \\\\ & |x| = |(x+y)-y| & \le & |x+y| + |y| \\ & |x| & \le & |x+y| + |y| \\ (2) & |x|-|y| & \le & |x+y| \\\\ \hline (1) \text{ and } (2) & ||x|-|y| | & \le & |x+y| \\ & |x+y| & \ge & ||x|-|y| | \\ \hline \end{array}\)

 

laugh

heureka  Jul 26, 2017
 #2
avatar+16 
+1

Thanks so much Heureka, legend!

AspiringActuary  Jul 26, 2017

19 Online Users

We use cookies to personalise content and ads, to provide social media features and to analyse our traffic. We also share information about your use of our site with our social media, advertising and analytics partners.  See details