document.write( "Question 962602: If a,b,c,d are in H.P., prove that a+d > b+c. \n" ); document.write( "
Algebra.Com's Answer #589826 by Edwin McCravy(20056)\"\" \"About 
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document.write( "PROOF:\r\n" );
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document.write( "This is not true unless we rule out negative numbers.  For here is a \r\n" );
document.write( "counter-example:\r\n" );
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document.write( "\"a=1%2F8\", \"b=1%2F5\", \"c=1%2F2\", \"d=1%2F%28-1%29\"  \r\n" );
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document.write( "are in H.P, because 8,5,2,-1 are in A.P. with common difference -3\r\n" );
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document.write( "yet \"a%2Bd=1%2F8-1=-7%2F8\" and \"b%2Bc=+7%2F10\" so a+d < b+c\r\n" );
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document.write( "So negative numbers cannot be allowed!\r\n" );
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document.write( "However it is true if a,b,c,d are all positive.  So you should point out \r\n" );
document.write( "to your teacher that the proposition is not true if you allow negative \r\n" );
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document.write( "So before we can prove the proposition, we must insert that requirement:\r\n" );
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\n" ); document.write( "If a,b,c,d are all positive and in H.P., prove that a+d > b+c.
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document.write( "Then there exists positive numbers in A.P., x,x+y,x+2y,x+3y where x > 0\r\n" );
document.write( "such that\r\n" );
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document.write( "\"a+=+1%2Fx\", \"b=1%2F%28x%2By%29\", \"c=1%2F%28x%2B2y%29\", \"d=1%2F%28x%2B3y%29\"\r\n" );
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document.write( "[Notice that although x is necessarily positive, y, the common difference, is\r\n" );
document.write( "NOT NECESSARILY positive!  However a+d and b+c are positive]\r\n" );
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document.write( "Then\r\n" );
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document.write( "\"%282x%2B3y%29%2F%28x%5E2%2B3xy%29\"\"%22%22%3E%22%22\"\"%282x%2B3y%29%2F%28x%5E2%2B3xy%2B2y%5E2%29\"\r\n" );
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document.write( "is true because the numerators are the same positive number and the\r\n" );
document.write( "denominator on the right is a larger positive number than the one\r\n" );
document.write( "on the left.  Therefore  \r\n" );
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document.write( "\"a%2Bd\"\"%22%22%3E%22%22\"\"b%2Bc\"\r\n" );
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document.write( "Edwin
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