document.write( "Question 710955: Prove (a+b+c)(ab+ac+bc)>= 9abc where a,b,c are positive.\r
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document.write( "I'm lost as to where to begin this proof \n" );
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Algebra.Com's Answer #437266 by Edwin McCravy(20056)![]() ![]() You can put this solution on YOUR website! \r\n" ); document.write( "Prove (a+b+c)(ab+ac+bc) ≧ 9abc where a,b,c are positive.\r\n" ); document.write( "\r\n" ); document.write( "The inequality is symmetric in a,b, and c, so\r\n" ); document.write( "without loss of generality, we can suppose a≦b≦c\r\n" ); document.write( "\r\n" ); document.write( "Then there exist non-negative numbers p, q, such that \r\n" ); document.write( "\r\n" ); document.write( "b = a+p and c = a+p+q\r\n" ); document.write( "\r\n" ); document.write( "Substitute those in the left side of the inequality:\r\n" ); document.write( "\r\n" ); document.write( "[a+(a+p)+(a+p+q)][a(a+p)+a(a+p+q)+(a+p)(a+p+q)]\r\n" ); document.write( "\r\n" ); document.write( "If you multiply that all the way out and collect terms (whew!),\r\n" ); document.write( "you will get this:\r\n" ); document.write( "\r\n" ); document.write( "9a³+18a²p+9a²q+11ap²+11apq+2aq²+2p³+3p²q+pq²\r\n" ); document.write( "\r\n" ); document.write( "Make the same substitutions in the right side of the inequality:\r\n" ); document.write( "\r\n" ); document.write( "9a(a+p)(a+p+q)\r\n" ); document.write( "\r\n" ); document.write( "If you multiply that all the way out and collect terms (not quite\r\n" ); document.write( "as tedious) you will get this:\r\n" ); document.write( "\r\n" ); document.write( "9a³+18a²p+9a²q+9ap²+9apq\r\n" ); document.write( "\r\n" ); document.write( "So the inequality we are to prove becomes\r\n" ); document.write( "\r\n" ); document.write( "9a³+18a²p+9a²q+11ap²+11apq+2aq²+2p³+3p²q+pq² ≧ 9a³+18a²p+9a²q+9ap²+9apq\r\n" ); document.write( "\r\n" ); document.write( "For contradiction assume the inequality can be <, that is, assume that\r\n" ); document.write( "for some non-negative p and q, this holds:\r\n" ); document.write( "\r\n" ); document.write( "9a³+18a²p+9a²q+11ap²+11apq+2aq²+2p³+3p²q+pq² < 9a³+18a²p+9a²q+9ap²+9apq\r\n" ); document.write( "\r\n" ); document.write( "Subtracting the right side from the left:\r\n" ); document.write( "\r\n" ); document.write( "2ap²+2apq+2aq²+2p³+3p²q+pq² < 0\r\n" ); document.write( "\r\n" ); document.write( "Since all the terms on the left are non-negative,\r\n" ); document.write( "this is clearly false, so < can never hold, thus \r\n" ); document.write( "we have reached a contradiction. ≧ always \r\n" ); document.write( "holds and the original inequality is proved true. QED\r\n" ); document.write( "\r\n" ); document.write( "Edwin\n" ); document.write( " |