document.write( "Question 711521: show a(b^2)+ b(c^2) + c(a^2) >= 3abc\r
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Algebra.Com's Answer #437451 by Edwin McCravy(20056)\"\" \"About 
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document.write( "ab² + bc² + ca² ≧ 3abc\r\n" );
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document.write( "Note:  the inequality does not hold if we allow negative \r\n" );
document.write( "numbers because if a=-1, b=5, and c=1,\r\n" );
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document.write( "-1·5² + 6·1² + 1·(-1)² ≧ 3·(-1)(1)\r\n" );
document.write( "  -25 + 6 + 1 ≧ -3\r\n" );
document.write( "          -18 ≧ -3\r\n" );
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document.write( "That is false.  So you must mean that a,b, and c are \r\n" );
document.write( "non-negative.\r\n" );
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document.write( "So I will assume that a,b,c are all non-negative.\r\n" );
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document.write( "The inequality is symmetric in a,b, and c.  Therefore\r\n" );
document.write( "without loss of generality we can assume a≦b≦c.\r\n" );
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document.write( "Then there exist non-negative numbers p and q such that\r\n" );
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document.write( "b = a+p, and c = a+p+q\r\n" );
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document.write( "Substituting those for b and c in the left side of the \r\n" );
document.write( "inequality gives:\r\n" );
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document.write( "a(a+p)² + (a+p)(a+p+q)² + (a+p+q)a²\r\n" );
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document.write( "Multiplying that out and collecting like terms gives:\r\n" );
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document.write( "3a³+6a²p+3a²q+4ap²+4apq+aq²+aq²+p³+2p²q+pq² \r\n" );
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document.write( "Making the same substitutions in the right side of the \r\n" );
document.write( "inequality gives:\r\n" );
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document.write( "3a(a+p)(a+p+q)\r\n" );
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document.write( "Multiplying that out and collecting like terms gives:\r\n" );
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document.write( "3a³+6a²p+3a²q+3ap²+3apq\r\n" );
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document.write( "So the inequality which we are to prove becomes:\r\n" );
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document.write( "3a³+6a²p+3a²q+4ap²+4apq+aq²+aq²+p³+2p²q+pq² ≧ 3a³+6a²p+3a²q+3ap²+3apq\r\n" );
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document.write( "For contradiction, assume that \r\n" );
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document.write( "3a³+6a²p+3a²q+4ap²+4apq+aq²+aq²+p³+2p²q+pq² < 3a³+6a²p+3a²q+3ap²+3apq\r\n" );
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document.write( "holds for some non-negative a, p, and q\r\n" );
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document.write( "Subtracting the right side from the left side we get\r\n" );
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document.write( "ap²+apq+2aq²+p³+2p²q+pq² < 0\r\n" );
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document.write( "But this can never hold because the left side is non-negative, \r\n" );
document.write( "so we have reached a contradiction.  Therefore the original \r\n" );
document.write( "inequality holds.\r\n" );
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document.write( "Edwin

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