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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!

 Jul 26, 2017
edited by Guest  Jul 26, 2017

Best Answer 

 #1
avatar+20850 
+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

 Jul 26, 2017
 #1
avatar+20850 
+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

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