document.write( "Question 984031: Prove that the geometric mean of two positive, unequal numbers is less than their arithmetic mean. \n" ); document.write( "
Algebra.Com's Answer #604838 by ikleyn(52792)\"\" \"About 
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\n" ); document.write( "\n" ); document.write( "Let  a  and  b  be two real positive unequal numbers.\r
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\n" ); document.write( "\n" ); document.write( "Their arithmetic mean is  \"%28a+%2B+b%29%2F2\".  Their geometric mean is  \"sqrt%28a%2Ab%29\".  We need to prove that \r
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\n" ); document.write( "\n" ); document.write( "\"sqrt%28a%2Ab%29\" < \"%28a+%2B+b%29%2F2\".\r
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\n" ); document.write( "\n" ); document.write( "Let us start with inequality\r
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\n" ); document.write( "\n" ); document.write( "\"%28sqrt%28a%29+-+sqrt%28b%29%29%5E2\" > 0.             (1)\r
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\n" ); document.write( "\n" ); document.write( "This inequality is true because the difference  \"sqrt%28a%29+-+sqrt%28b%29\"  is not zero and the square of a non-zero real number is positive.  Now,  expand  (1)  as \r
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\n" ); document.write( "\n" ); document.write( "\"%28sqrt%28a%29+-+sqrt%28b%29%29%5E2\" = \"%28sqrt%28a%29%29%5E2\" - \"2%2Asqrt%28a%29%2Asqrt%28b%29\" + \"%28sqrt%28b%29%29%5E2\" = \"a\" - \"2%2Asqrt%28a%29%2Asqrt%28b%29\" + \"b\".\r
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\n" ); document.write( "\n" ); document.write( "Thus the original inequality  (1)  is equivalent to \r
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\n" ); document.write( "\n" ); document.write( "\"a\" - \"2%2Asqrt%28a%29%2Asqrt%28b%29\" + \"b\" > \"0\",     or\r
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\n" ); document.write( "\n" ); document.write( "\"a\" + \"b\" > \"2%2Asqrt%28a%29%2Asqrt%28b%29\",     or\r
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\n" ); document.write( "\n" ); document.write( "\"%28a+%2B+b%29%2F2\" > \"sqrt%28a%2Ab%29\". \r
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\n" ); document.write( "\n" ); document.write( "It is exactly what has to be proved. \r
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