document.write( "Question 1015301: If a,b,c are in G.P. Then prove that (a^2+ab+b^2)/(bc+ca+ab)=(b+a)/(c+b) \n" ); document.write( "

Algebra.Com's Answer #631662 by Edwin McCravy(20055) $\"\"$ $\"About$

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document.write( "We are given that a,b,c are in G.P.  Therefore:\r\n" );
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document.write( "or\r\n" );
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document.write( "or upon cross-multiplying:\r\n" );
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document.write( "(1)   \r\n" );
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document.write( "When the conclusion is more complicated than the premise,\r\n" );
document.write( "the best plan is to do an INDIRECT proof.  We will begin\r\n" );
document.write( "by assuming the premise true, that is, that a,b,c are in \r\n" );
document.write( "G.P., but that the conclusion: \r\n" );
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document.write( "is false. \r\n" );
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document.write( "That is we start by assuming the conclusion is false, i.e., that:\r\n" );
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document.write( "Let's use (1) above to substitute ac for b² in the numerator,\r\n" );
document.write( "and b² for ca, which is the same as ac,  in the denominator\r\n" );
document.write( "on the left:\r\n" );
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document.write( "Factor a out of the numerator and b out of the denominator:\r\n" );
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document.write( "We can cancel (a+b+c) and (c+b+a) on the left and get\r\n" );
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document.write( "The inequality will still hold if we cross-multiply:\r\n" );
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document.write( "The inequality will still hold if we subtract ab from both\r\n" );
document.write( "sides:\r\n" );
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document.write( "This contradicts (1).\r\n" );
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document.write( "Therefore the assumption \r\n" );
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document.write( "is false and therefore:\r\n" );
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document.write( "Edwin

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