document.write( "Question 1209502: Let be three positive numbers such that:\r
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\n" ); document.write( "\n" ); document.write( "x + y + z + \frac{1}{xyz} > 4.\r
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Algebra.Com's Answer #849024 by ikleyn(52781) $\"\"$ $\"About$

You can put this solution on YOUR website!
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\n" ); document.write( "Let x, y, z be three positive numbers such that:\r
\n" ); document.write( "\n" ); document.write( "x^2 + y^2 + z^2 = 2(xy + xz + yz).\r
\n" ); document.write( "\n" ); document.write( "Prove that x + y + z + 1/(xyz) > 4.
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\n" ); document.write( "\n" ); document.write( "                The solution in the post by @CPhill is    $\"highlight%28highlight%28INCORRECT%29%29\"$ .\r
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\n" ); document.write( "\n" ); document.write( "        Indeed, he rearranges the given equation   x^2 + y^2 + z^2 = 2(xy + xz + yz)\r
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\n" ); document.write( "\n" ); document.write( "        to equation   x^2 + y^2 + z^2 - 2xy - 2xz - 2yz = 0,   which is right.\r
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\n" ); document.write( "\n" ); document.write( "        But then he \" factors \" the left-hand side \r
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\n" ); document.write( "\n" ); document.write( "                (x-y)^2 + (y-z)^2 + (z-x)^2 = 0,\r
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\n" ); document.write( "\n" ); document.write( "        which is FATALLY WRONG.\r
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\n" ); document.write( "\n" ); document.write( "        Hence, everything what @CPhill derives further from the \" factored \" form is incorrect \r
\n" ); document.write( "\n" ); document.write( "        and does not have the power of logical deducing.\r
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\n" ); document.write( "\n" ); document.write( "                 +-----------------------------------------------------------------+
\n" ); document.write( "                     Nevertheless, the statement of the problem is correct,
\n" ); document.write( "                                 and below I developed another solution.
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document.write( "We start from equation\r\n" );
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document.write( "    x^2 + y^2 + z^2 = 2xy + 2xz + 2yz,    (1)\r\n" );
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document.write( "which is given.   Add 2xy + 2xz + 2yz to both side.  You will get\r\n" );
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document.write( "    x^2 + y^ + z^2 + 2xy + 2xz + 2yz = 4xy + 4xz + 4 yz.    (2)\r\n" );
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document.write( "Left side of the last equation is  (x+y+z)^2,  so it can be re-written in the form\r\n" );
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document.write( "    (x + y + z)^2 = 4(xy + xz + yz).    (3)\r\n" );
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document.write( "Now apply AM-GM (Arithmetic mean - Geometric mean) inequality, which says\r\n" );
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document.write( "     >=     (4)\r\n" );
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document.write( "for any three positive numbers a, b, c.\r\n" );
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document.write( "Therefore, for the right side of equation (3) we have this inequality\r\n" );
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document.write( "    4(xy + xz + yz) =  >=  = .    (5)\r\n" );
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document.write( "So, combining (3) and (5), we get\r\n" );
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document.write( "    (x + y + z)^2 >= .    (6)\r\n" );
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document.write( "Now take square root of both sides of (6) and get\r\n" );
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document.write( "    x + y + z >= .     (7)\r\n" );
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document.write( "Now, to prove the inequality as it is given in the condition, add  to both sides of (7).\r\n" );
document.write( "You will get\r\n" );
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document.write( "    x + y + z +  >=  + .    (8)\r\n" );
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document.write( "Consider function   + .\r\n" );
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document.write( "This function is easy to analyze for its minimum.\r\n" );
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document.write( "Such an analysis can be done using standard Calculus procedure or graphically.\r\n" );
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document.write( "The plot of this function is shown in Figure under this link\r\n" );
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document.write( "https://www.desmos.com/calculator/zw8xlstrvb\r\n" );
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document.write( "The minimum is  4.45566.  For us, it gives the information that right side of (8) is always greater than 4.\r\n" );
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document.write( "So, the inequality\r\n" );
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document.write( "    x + y + z +  > 4\r\n" );
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document.write( "is proved.\r\n" );
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