document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=1209502>Question 1209502</A>: <I><SPAN STYLE=\"word-break: break-all;\"> Let  be three positive numbers such that:\r<BR>\n" );
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document.write( "x^2 + y^2 + z^2 = 2(xy + xz + yz).\r<BR>\n" );
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document.write( "x + y + z + \frac{1}{xyz} > 4.\r<BR>\n" );
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document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #848934 by </B><A HREF=/tutors/aboutme.mpl?userid=CPhill><B>CPhill(1959)</B></A><img src=\"/images/stars/stars-12.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=CPhill><IMG SRC=http://www.algebra.com/images/aboutme-small.gif BORDER=0 alt=\"About Me\" VALIGN=MIDDLE></A>&nbsp;<IMG SRC=http://www.algebra.com/cgi-bin/show-face.mpl?user=CPhill><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=848934\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> Certainly, let's prove the given inequality.\r<BR>\n" );
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document.write( "**1. Utilize the Given Condition**\r<BR>\n" );
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document.write( "* We are given: x² + y² + z² = 2(xy + xz + yz) \r<BR>\n" );
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document.write( "* Rearrange the equation: <BR>\n" );
document.write( "   x² + y² + z² - 2xy - 2xz - 2yz = 0\r<BR>\n" );
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document.write( "* Factor the left-hand side:<BR>\n" );
document.write( "   (x - y)² + (y - z)² + (z - x)² = 0\r<BR>\n" );
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document.write( "* Since the squares of real numbers are always non-negative, the only way for the sum of three squares to be zero is if each individual square is zero. \r<BR>\n" );
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document.write( "* Therefore:<BR>\n" );
document.write( "   x - y = 0 <BR>\n" );
document.write( "   y - z = 0<BR>\n" );
document.write( "   z - x = 0\r<BR>\n" );
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document.write( "* This implies that:<BR>\n" );
document.write( "   x = y = z\r<BR>\n" );
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document.write( "**2. Substitute and Simplify**\r<BR>\n" );
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document.write( "* Let x = y = z = k (where k is a positive number)\r<BR>\n" );
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document.write( "* The inequality to prove becomes:<BR>\n" );
document.write( "   k + k + k + 1/(k * k * k) > 4\r<BR>\n" );
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document.write( "   3k + 1/k³ > 4\r<BR>\n" );
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document.write( "**3. Prove the Inequality**\r<BR>\n" );
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document.write( "* **Method 1: AM-GM Inequality**\r<BR>\n" );
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document.write( "   * Apply the AM-GM inequality to the numbers 3k and 1/k³:<BR>\n" );
document.write( "      (3k + 1/k³) / 2 ≥ √(3k * 1/k³) <BR>\n" );
document.write( "      (3k + 1/k³) / 2 ≥ √3 <BR>\n" );
document.write( "      3k + 1/k³ ≥ 2√3\r<BR>\n" );
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document.write( "   * Since 2√3 > 4, we have:<BR>\n" );
document.write( "      3k + 1/k³ > 4 \r<BR>\n" );
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document.write( "* **Method 2: Algebraic Manipulation**\r<BR>\n" );
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document.write( "   * Multiply both sides of the inequality by k³ (since k is positive, this does not change the direction of the inequality):<BR>\n" );
document.write( "      3k⁴ + 1 > 4k³\r<BR>\n" );
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document.write( "   * Rearrange:<BR>\n" );
document.write( "      3k⁴ - 4k³ + 1 > 0\r<BR>\n" );
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document.write( "   * Factor:<BR>\n" );
document.write( "      (k - 1)³ (3k + 1) > 0\r<BR>\n" );
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document.write( "   * Since k is positive, (3k + 1) is always positive. \r<BR>\n" );
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document.write( "   * Therefore, for the inequality to hold, (k - 1)³ must be positive. This implies k > 1.\r<BR>\n" );
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document.write( "   * Since x = y = z = k, we have x > 1, y > 1, and z > 1. \r<BR>\n" );
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document.write( "   * Now, consider the expression:<BR>\n" );
document.write( "      x + y + z + 1/(xyz) \r<BR>\n" );
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document.write( "   * Since x, y, and z are all greater than 1, 1/(xyz) will be less than 1.\r<BR>\n" );
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document.write( "   * Therefore, x + y + z + 1/(xyz) > x + y + z > 3 > 4\r<BR>\n" );
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document.write( "**Conclusion**\r<BR>\n" );
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document.write( "We have proven that if x², y², and z² satisfy the given condition, then x + y + z + 1/(xyz) > 4.<BR>\n" );
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