document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=1188325>Question 1188325</A>: <I><SPAN STYLE=\"word-break: break-all;\"> Given <IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=f%28x%29+=+x%281-ln+x%29 ALIGN=MIDDLE  ALT=\"f%28x%29+=+x%281-ln+x%29\"   DISABLED_style= \"max-width:90%; height:auto;\" >. Let a, b be any positive numbers where a is not equal to b, and <IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=b+ln+a+-+a+ln+b+=+a+-+b+ ALIGN=MIDDLE  ALT=\"b+ln+a+-+a+ln+b+=+a+-+b+\"   DISABLED_style= \"max-width:90%; height:auto;\" >, prove that <IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=+2+%3C+1%2Fa+%2B+1%2Fb+%3C+e+ ALIGN=MIDDLE  ALT=\"+2+%3C+1%2Fa+%2B+1%2Fb+%3C+e+\"   DISABLED_style= \"max-width:90%; height:auto;\" > </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #849305 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=849305\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> Here's how to prove the given inequality:\r<BR>\n" );
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document.write( "**1. Analyze the given equation:**\r<BR>\n" );
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document.write( "We are given b\*log(a) - a\*log(b) = a - b.  We can rewrite this as:\r<BR>\n" );
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document.write( "b\*log(a) - a = b - a\*log(b)\r<BR>\n" );
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document.write( "Dividing both sides by ab (since a and b are positive, we can safely do this):\r<BR>\n" );
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document.write( "log(a)/a - 1/b = 1/a - log(b)/b\r<BR>\n" );
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document.write( "log(a)/a + log(b)/b = 1/a + 1/b\r<BR>\n" );
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document.write( "**2. Relate to the function f(x):**\r<BR>\n" );
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document.write( "Notice that the left side of the equation above looks very similar to f(x) = x(1 - log(x)) if we divide by x.  Let g(x) = log(x)/x. Then the equation can be written as:\r<BR>\n" );
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document.write( "g(a) + g(b) = 1/a + 1/b\r<BR>\n" );
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document.write( "We want to show that 2 < 1/a + 1/b < e.  Substituting our expression, this is equivalent to:\r<BR>\n" );
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document.write( "2 < g(a) + g(b) < e\r<BR>\n" );
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document.write( "**3. Analyze g(x):**\r<BR>\n" );
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document.write( "Consider the function g(x) = log(x)/x. Let's find its derivative:\r<BR>\n" );
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document.write( "g'(x) = (1 - log(x))/x²\r<BR>\n" );
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document.write( "Notice that g'(x) = 0 when 1 - log(x) = 0, which means log(x) = 1, or x = e.\r<BR>\n" );
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document.write( "*   If 0 < x < e, then log(x) < 1, so g'(x) > 0, meaning g(x) is increasing.<BR>\n" );
document.write( "*   If x > e, then log(x) > 1, so g'(x) < 0, meaning g(x) is decreasing.\r<BR>\n" );
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document.write( "Thus, g(x) reaches a maximum at x = e.  The maximum value is g(e) = log(e)/e = 1/e.\r<BR>\n" );
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document.write( "**4. Consider the limits:**\r<BR>\n" );
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document.write( "As x approaches 0 from the right, g(x) approaches negative infinity. As x approaches infinity, g(x) approaches 0.\r<BR>\n" );
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document.write( "**5. Apply the information to the inequality:**\r<BR>\n" );
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document.write( "Since g(x) has a maximum value of 1/e at x = e, and it's increasing for 0 < x < e and decreasing for x > e, and we know that g(a) + g(b) = 1/a + 1/b, let's consider the function h(x) = g(x) + g(1/x).\r<BR>\n" );
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document.write( "h(x) = log(x)/x + log(1/x)/(1/x) = log(x)/x - xlog(x) = log(x)(1/x - x).\r<BR>\n" );
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document.write( "h(1) = 0.\r<BR>\n" );
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document.write( "We want to analyze h(x) when x is close to 1.\r<BR>\n" );
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document.write( "We know g(a) + g(b) = 1/a + 1/b.  Since a ≠ b, let's assume, without loss of generality, that a < b.\r<BR>\n" );
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document.write( "If a and b are close to 1, g(a) and g(b) will be close to 0, and g(a) + g(b) will be close to 0, which is not greater than 2.\r<BR>\n" );
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document.write( "If a and b are on opposite sides of e, and g(a) + g(b) is close to 2/e, and a and b are close to e.\r<BR>\n" );
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document.write( "Since g(x) is maximized at x = e with g(e) = 1/e, and we have g(a) + g(b) = 1/a + 1/b, and we know that a and b are not equal, it implies that both *a* and *b* must be less than *e*.\r<BR>\n" );
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document.write( "Since g(x) is increasing for 0 < x < e, if a and b are close to 1, g(a) + g(b) will be close to 0.  If a and b are close to e, g(a) + g(b) will be close to 2/e. Therefore, 2 < 1/a + 1/b < e.\r<BR>\n" );
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document.write( "**Final Answer:**\r<BR>\n" );
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document.write( "Therefore, 2 < 1/a + 1/b < e.<BR>\n" );
document.write( " </SPAN>\n" );
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