document.write( "Question 982143: prove that the geometric mean of two positive, unequal numbers is less than their arithmetic mean. \n" ); document.write( "
Algebra.Com's Answer #603005 by ikleyn(52798)\"\" \"About 
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\n" ); document.write( "\n" ); document.write( "Let  a  and  b  be two positive, unequal real numbers. \r
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\n" ); document.write( "\n" ); document.write( "Then their arithmetic mean is  \"%28a+%2B+b%29%2F2\"  and geometric mean is  \"sqrt%28a%2Ab%29\".\r
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\n" ); document.write( "\n" ); document.write( "We need to prove that \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( "Let us start with this inequality\r
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\n" ); document.write( "\n" ); document.write( "\"%28sqrt%28a%29+-+sqrt%28b%29%29%5E2\" > 0,\r
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\n" ); document.write( "\n" ); document.write( "which is always true for any real unequal numbers  a  and  b,  because the left part is the square of the real number  \"sqrt%28a%29+-+sqrt%28b%29\",  which is unequal to zero.\r
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\n" ); document.write( "\n" ); document.write( "Rewrite this inequality in this way step by step:\r
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\n" ); document.write( "\n" ); document.write( "\"a+-+2sqrt%28a%29%2Asqrt%28b%29+%2B+b\" > \"0\",         (apply the formula of the square of a difference)\r
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\n" ); document.write( "\n" ); document.write( "\"a+%2B+b\" > \"2sqrt%28a%29%2Asqrt%28b%29\",             (move the term  \"-+2sqrt%28a%29%2Asqrt%28b%29\"  from the left side to the right with the opposite sign)\r
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\n" ); document.write( "\n" ); document.write( "\"%28a%2Bb%29%2F2\" > \"sqrt%28a%2Ab%29\". \r
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\n" ); document.write( "\n" ); document.write( "The last inequality is exactly what you have to prove. \r
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