document.write( "Question 711123: If a, b, c are positive real numbers prove that
\n" ); document.write( "(a + b + c)(1/a + 1/b + 1/c) >= 9\r
\n" ); document.write( "\n" ); document.write( "I'm lost as to how to begin this proof
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Algebra.Com's Answer #437332 by KMST(5328)\"\" \"About 
You can put this solution on YOUR website!
First you multiply and simplify.
\n" ); document.write( "I would also rearrange as shown below.
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\n" ); document.write( "Next we have to show that each expression in brackets is equal or greater than 2.
\n" ); document.write( "All of them are of the form
\n" ); document.write( "\"x%2Fy%2By%2Fx\"
\n" ); document.write( "We could probe it with \"x%2Fy%2By%2Fx\" or with \"a%2Fb%2Bb%2Fa\".
\n" ); document.write( "\"x%2Fy%2By%2Fx=%28x%5E2%2By%5E2%29%2Fxy\"
\n" ); document.write( "We want to prove that \"%28x%5E2%2By%5E2%29%2Fxy%3E=2\"
\n" ); document.write( "\"%28x%5E2%2By%5E2%29-2xy=%28x-y%29%5E2\"
\n" ); document.write( "so \"%28x%5E2%2By%5E2%29-2xy%3E=0\" --> \"%28x%5E2%2By%5E2%29%3E=2xy\"
\n" ); document.write( "And since x and y are positive xy is a positive number we can use to divide both sides of the inequality to find that
\n" ); document.write( "\"%28x%5E2%2By%5E2%29%2Fxy%3E=2xy%2Fxy\" --> \"%28x%5E2%2By%5E2%29%2Fxy%3E=2\" --> \"x%2Fy%2By%2Fx%3E=2\"
\n" ); document.write( "Then
\n" ); document.write( "\"a%2Fb%2Bb%2Fa%3E=2\" ,
\n" ); document.write( "\"a%2Fc%2Bc%2Fa%3E=2\" and
\n" ); document.write( "\"b%2Fc%2Bc%2Fb%3E=2\"
\n" ); document.write( "So
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\n" ); document.write( "\"%28a+%2B+b+%2B+c%29%281%2Fa+%2B+1%2Fb+%2B+1%2Fc%29%3E=9\"
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