document.write( "Question 643963: prove that abs(x-y)>abs(x)-abs(y) \n" ); document.write( "
Algebra.Com's Answer #404763 by Edwin McCravy(20054)\"\" \"About 
You can put this solution on YOUR website!
abs(x-y)>abs(x)-abs(y)
\n" ); document.write( "
\r\n" );
document.write( "The > should be ≧ because they are often equal, e.g., |3-2| = |3|-|2|,\r\n" );
document.write( "so I will change your problem to\r\n" );
document.write( "\r\n" );
document.write( "Prove that |x-y| ≧ |x|-|y|\r\n" );
document.write( "\r\n" );
document.write( "\r\n" );
document.write( "\r\n" );
document.write( "       xy ≦ |xy|               because they are equal if either x or y is 0 or if\r\n" );
document.write( "                               they are both positive or both negative, and if\r\n" );
document.write( "                               they have opposite signs the left side is\r\n" );
document.write( "                               negative and the right side is positive, so the\r\n" );
document.write( "                               inequality holds in all cases.\r\n" );
document.write( "\r\n" );
document.write( "     -2xy ≧ -2|xy|             We multiplied both sides by -2, which reversed\r\n" );
document.write( "                               the inequality.\r\n" );
document.write( "\r\n" );
document.write( "x²-2xy+y² ≧ x²-2|xy|+y²        We added x²+y² to both sides.\r\n" );
document.write( "\r\n" );
document.write( "x²-2xy+y² ≧ |x|²-2|x||y|+|y|²  On the right, we replaced x² by |x|², y²\r\n" );
document.write( "                               by |y|² and |xy| by |x||y|\r\n" );
document.write( "\r\n" );
document.write( "   (x-y)² ≧ (|x|-|y|)²         We factored both sides.\r\n" );
document.write( "\r\n" );
document.write( "    |x-y| ≧ |x|-|y|            We took non-negative square roots of both sides.    \r\n" );
document.write( "\r\n" );
document.write( " Edwin
\n" ); document.write( "
\n" );