+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!
+26393 Using triangle inequity prove |x+y|>=||x|-|y||
\(\begin{array}{|rcll|} \hline |x+y| & \stackrel{?} \ge & ||x|-|y|| \\ \hline \end{array}\)
\(\begin{array}{|lrcll|} \hline &|y| = |(x+y)-x| & \le & |x+y| + |x| \\ & |y| & \le & |x+y| + |x| \\ (1) & |y|-|x| & \le & |x+y| \\\\ & |x| = |(x+y)-y| & \le & |x+y| + |y| \\ & |x| & \le & |x+y| + |y| \\ (2) & |x|-|y| & \le & |x+y| \\\\ \hline (1) \text{ and } (2) & ||x|-|y| | & \le & |x+y| \\ & |x+y| & \ge & ||x|-|y| | \\ \hline \end{array}\)
+26393 Using triangle inequity prove |x+y|>=||x|-|y||
\(\begin{array}{|rcll|} \hline |x+y| & \stackrel{?} \ge & ||x|-|y|| \\ \hline \end{array}\)
\(\begin{array}{|lrcll|} \hline &|y| = |(x+y)-x| & \le & |x+y| + |x| \\ & |y| & \le & |x+y| + |x| \\ (1) & |y|-|x| & \le & |x+y| \\\\ & |x| = |(x+y)-y| & \le & |x+y| + |y| \\ & |x| & \le & |x+y| + |y| \\ (2) & |x|-|y| & \le & |x+y| \\\\ \hline (1) \text{ and } (2) & ||x|-|y| | & \le & |x+y| \\ & |x+y| & \ge & ||x|-|y| | \\ \hline \end{array}\)
