document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=1209918>Question 1209918</A>: <I><SPAN STYLE=\"word-break: break-all;\"> Let x_1, x_2, \dots, x_n be real numbers.  If<BR>\n" );
document.write( "x_1 + 2x_2 + \dots + nx_n = 1,<BR>\n" );
document.write( "then find the minimum value of x_1^2/1 + x_2^2/2 + \dots + x_n^2/n, in terms of n.<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 #851058 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=851058\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> Let $a_k = kx_k$ and $b_k = 1/k$.<BR>\n" );
document.write( "Then the given condition is<BR>\n" );
document.write( "$$\sum_{k=1}^n a_k = \sum_{k=1}^n kx_k = 1$$<BR>\n" );
document.write( "We want to find the minimum value of<BR>\n" );
document.write( "$$S = \sum_{k=1}^n \frac{x_k^2}{k} = \sum_{k=1}^n \frac{a_k^2}{k^3}$$\r<BR>\n" );
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document.write( "By the Cauchy-Schwarz inequality, we have<BR>\n" );
document.write( "$$\left( \sum_{k=1}^n a_k \right)^2 \le \left( \sum_{k=1}^n \frac{a_k^2}{k^3} \right) \left( \sum_{k=1}^n k^3 \right)$$<BR>\n" );
document.write( "Since $\sum_{k=1}^n a_k = 1$, we have<BR>\n" );
document.write( "$$1 \le \left( \sum_{k=1}^n \frac{a_k^2}{k^3} \right) \left( \sum_{k=1}^n k^3 \right)$$<BR>\n" );
document.write( "$$\sum_{k=1}^n \frac{a_k^2}{k^3} \ge \frac{1}{\sum_{k=1}^n k^3}$$<BR>\n" );
document.write( "Therefore,<BR>\n" );
document.write( "$$S \ge \frac{1}{\sum_{k=1}^n k^3}$$\r<BR>\n" );
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document.write( "The minimum value of $\sum_{k=1}^n \frac{x_k^2}{k}$ is $\frac{1}{\sum_{k=1}^n k^3}$.<BR>\n" );
document.write( "We know that $\sum_{k=1}^n k^3 = \left( \frac{n(n+1)}{2} \right)^2$.<BR>\n" );
document.write( "So the minimum value is<BR>\n" );
document.write( "$$\frac{1}{\left( \frac{n(n+1)}{2} \right)^2} = \frac{4}{n^2(n+1)^2}$$\r<BR>\n" );
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document.write( "To achieve equality, we need<BR>\n" );
document.write( "$$\frac{a_1}{1^3} = \frac{a_2}{2^3} = \dots = \frac{a_n}{n^3} = c$$<BR>\n" );
document.write( "where $c$ is a constant.<BR>\n" );
document.write( "Then $a_k = ck^3$, so $kx_k = ck^3$, which means $x_k = ck^2$.<BR>\n" );
document.write( "Substituting into the given condition, we have<BR>\n" );
document.write( "$$\sum_{k=1}^n kx_k = \sum_{k=1}^n k(ck^2) = c \sum_{k=1}^n k^3 = 1$$<BR>\n" );
document.write( "$$c = \frac{1}{\sum_{k=1}^n k^3}$$<BR>\n" );
document.write( "Then $x_k = \frac{k^2}{\sum_{k=1}^n k^3}$.\r<BR>\n" );
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document.write( "Therefore, the minimum value of $\sum_{k=1}^n \frac{x_k^2}{k}$ is $\frac{1}{\sum_{k=1}^n k^3} = \frac{4}{n^2(n+1)^2}$.\r<BR>\n" );
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document.write( "Final Answer: The final answer is $\boxed{\frac{4}{n^2(n+1)^2}}$<BR>\n" );
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