document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=464215>Question 464215</A>: <I><SPAN STYLE=\"word-break: break-all;\"> prove that if a,b,c are real, the roots of (1/x+a) +(1/x+b) +(1/x+c) =3/x are also real </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #317992 by </B><A HREF=/tutors/aboutme.mpl?userid=richard1234><B>richard1234(7193)</B></A><img src=\"/images/stars/stars-13.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=richard1234><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=richard1234><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=317992\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> I presume you mean\r<BR>\n" );
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document.write( "<IMG STYLE=\"max-width:80%;\" SRC=\"/cgi-bin/rendertex.cgi?\LARGE \frac{1}{x+a} + \frac{1}{x+b} + \frac{1}{x+c} = \frac{3}{x}\">\r<BR>\n" );
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document.write( "in which we want to prove that all roots x that satisfy are real. Suppose we rewrite 3/x as 1/x + 1/x + 1/x:\r<BR>\n" );
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document.write( "<IMG STYLE=\"max-width:80%;\" SRC=\"/cgi-bin/rendertex.cgi?\LARGE \frac{1}{x+a} + \frac{1}{x+b} + \frac{1}{x+c} = \frac{1}{x} + \frac{1}{x} + \frac{1}{x}\">\r<BR>\n" );
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document.write( "<IMG STYLE=\"max-width:80%;\" SRC=\"/cgi-bin/rendertex.cgi?\LARGE (\frac{1}{x+a} - \frac{1}{x}) + (\frac{1}{x+b} - \frac{1}{x}) + (\frac{1}{x+c} - \frac{1}{x}) = 0\">\r<BR>\n" );
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document.write( "<IMG STYLE=\"max-width:80%;\" SRC=\"/cgi-bin/rendertex.cgi?\LARGE \frac{a}{x(x+a)} + \frac{b}{x(x+b)} + \frac{c}{x(x+c)} = 0\"> (combined fractions, then multiplied by -1)\r<BR>\n" );
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document.write( "Provided <IMG STYLE=\"max-width:80%;\" SRC=\"/cgi-bin/rendertex.cgi?x \not= 0\"> we can multiply both sides by x to cancel the x terms out. After this, we can multiply both sides by (x+a)(x+b)(x+c):\r<BR>\n" );
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document.write( "<IMG STYLE=\"max-width:80%;\" SRC=\"/cgi-bin/rendertex.cgi?\LARGE a(x+b)(x+c) + b(x+a)(x+c) + c(x+a)(x+b) = 0\">\r<BR>\n" );
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document.write( "This will result in a nice second-degree polynomial:\r<BR>\n" );
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document.write( "<IMG STYLE=\"max-width:80%;\" SRC=\"/cgi-bin/rendertex.cgi?\LARGE (a+b+c)x^2 + 2(ab + bc + ca)x + 3abc = 0\">\r<BR>\n" );
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document.write( "All that is left to do is prove that the discriminant of this quadratic is nonnegative, in other words,\r<BR>\n" );
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document.write( "<IMG STYLE=\"max-width:80%;\" SRC=\"/cgi-bin/rendertex.cgi?\LARGE 4(ab + bc + ca)^2 - 12(a+b+c)abc \ge 0 \">\r<BR>\n" );
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document.write( "<IMG STYLE=\"max-width:80%;\" SRC=\"/cgi-bin/rendertex.cgi?\LARGE (ab + bc + ca)^2 - 3(a+b+c)abc \ge 0\">\r<BR>\n" );
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document.write( "<IMG STYLE=\"max-width:80%;\" SRC=\"/cgi-bin/rendertex.cgi?\LARGE a^2b^2 + b^2c^2 + c^2a^2 - a^2bc - ab^2c - abc^2 \ge 0\">\r<BR>\n" );
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document.write( "<IMG STYLE=\"max-width:80%;\" SRC=\"/cgi-bin/rendertex.cgi?\LARGE a^2b^2 + b^2c^2 + c^2a^2 \ge a^2bc + ab^2c + abc^2 (*)\">\r<BR>\n" );
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document.write( "Note that this is true, since we can apply the Cauchy-Schwarz inequality, which tells us that\r<BR>\n" );
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document.write( "<IMG STYLE=\"max-width:80%;\" SRC=\"/cgi-bin/rendertex.cgi?\LARGE \sum_{a,b,c} a^2b^2\sum_{a,b,c} b^2c^2 \ge (\sum_{a,b,c} ab^2c)^2\"> (shorthand for cyclic sum)\r<BR>\n" );
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document.write( "<IMG STYLE=\"max-width:80%;\" SRC=\"/cgi-bin/rendertex.cgi?\LARGE a^2b^2 + b^2c^2 + c^2a^2 \ge ab^2c + abc^2 + a^2bc (*)\">\r<BR>\n" );
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document.write( "Hence, this implies that the discriminant is positive and the roots of the quadratic are both real. However I do not quite remember if the Cauchy-Schwarz inequality can apply for negative a,b,c (I'm pretty sure it does though, unlike AM-GM), but if it doesn't you might be able to generalize for negative a,b,c.<BR>\n" );
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