Algebra.Com's Answer #847416 by ikleyn(52810)![\"\"](\"/images/stars/stars-13.gif\")  
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document.write( "For 0 < a < b, let h be defined by\r
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document.write( "1/h = (1/2)[(1/a) + (1/b)].\r
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document.write( "Show that a < h < b. \r
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document.write( "Note: The number h is called the harmonic mean of a and b.
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document.write( "Let' simplify\r\n" );
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document.write( "  =  =  .\r\n" );
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document.write( "Since  =  , it implies that h =  .\r\n" );
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document.write( "Now we want to prove that \r\n" );
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document.write( " a < < b. (*)\r\n" );
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document.write( "So, your starting inequality is \r\n" );
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document.write( " a < b. (1)\r\n" );
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document.write( "In (1), multiply both sides by positive number \"a\". \r\n" );
document.write( "You will get an equivalent inequality\r\n" );
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document.write( " a^2 < ab. (2)\r\n" );
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document.write( "Next step, in (1), multiply both sides by positive number \"b\". \r\n" );
document.write( "You will get an equivalent inequality\r\n" );
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document.write( " ab < b^2. (3)\r\n" );
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document.write( "From (2) and (3), you have this compound inequality\r\n" );
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document.write( " a^2 < ab < b^2. (4)\r\n" );
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document.write( "Add ab to all three terms in inequality (4). You will get an equivalent inequality \r\n" );
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document.write( " a^2 + ab < ab + ab < b^2 + ab. (5)\r\n" );
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document.write( "Rewrite it equivalently this way\r\n" );
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document.write( " a*(a+b) < 2ab < b*(a+b). (6)\r\n" );
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document.write( "In (6), divide all three sides by positive real number a+b. You will get an equivalent inequality\r\n" );
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document.write( " a < < b. (7)\r\n" );
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document.write( "Compare it with (*) : inequality (7) is what you want to prove.\r\n" );
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document.write( "Solved.\r
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