Let \(x\) and \(y\) be positive real numbers such that \(mx+y=1.\) Find the positive \(m\) such that the minimum of \(\left( 1 + \frac{1}{x} \right)\left( 1 + \frac{m}{y} \right)\) is 81.

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Let \(x\) and \(y\) be positive real numbers such that \(mx+y=1.\) Find the positive \(m\) such that the minimum of \(\left( 1 + \frac{1}{x} \right)\left( 1 + \frac{m}{y} \right)\) is 81.

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solve first for y and sub in the other equation to get only x. then rework equation to simplify. then you can differentiate, set equal to 81 and solve
can you use AM-GM here because when you add the terms in brackets you get \(2 + \frac{1}{xy}\) so you end up solving something like \(\frac{1}{2}(2 + \frac{1}{xy})^2 > 81\) subject to \(y + mx = 1\) and funnily enough 81 is a square number
i know *no* pure maths.

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so could you say \(\frac{1}{\sqrt{2}}\sqrt{81} \lt \frac{1}{\sqrt{2}}\sqrt{\left( 1 + \frac{1}{x} \right)\left( 1 + \frac{m}{y} \right)}\lt \frac{1}{2} \left( 1 + \frac{1}{x} + 1 + \frac{m}{y} \right)\) and go from there?!?!
\[\frac{1}{2}(81) \lt \frac{1}{4} \left( 2 + \frac{y+mx}{xy} \right)^2\] \[\left( 2 + \frac{1}{xy} \right) \gt 9 \sqrt{2} \]
@freckles @Loser66 @ganeshie8 no idea how to differentiate - this is a from a precalc class. Is there a simpler way to solve this problem? Not sure what @IrishBoy123 is doing. Thanks!
\[ \left(1+\frac{1}{x}\right)\left(1+\frac{m}{y}\right)=1+\frac{1}{x}+\frac{m}{y}+\frac{m}{xy}\\ \frac{1}{4}\left(1+\frac{1}{x}+\frac{m}{y}+\frac{m}{xy}\right)\geq\sqrt[4]{\frac{m^2}{x^2y^2}}=\sqrt{\frac{|m|}{xy}}\\ \frac{1}{4}\left(1+\frac{1}{x}+\frac{m}{y}+\frac{m}{xy}\right)=\sqrt{\frac{|m|}{xy}}\quad \text{AM-GM}\\ 1+\frac{1}{x}+\frac{m}{y}+\frac{m}{xy}=4\sqrt{\frac{|m|}{xy}}=81\\ 4\sqrt{\frac{|m|}{xy}}=81\\\\ \frac{|m|}{xy}=\frac{6561}{16}\\ mx+y=1\implies m<0\\ \frac{m}{xy}=\frac{-6561}{16}=\frac{1}{x}=\frac{m}{y}=1!?\\ \] Furthermore, Mathematica shows that there is no unique m that satisfies the requirement. For every different x and y there is a different but unique m that satisfies it.
\[\begin{align} &\left(1+\frac{1}{x}\right)\left(1+\frac{m}{y}\right)\\ &=1+\frac{1}{x}+\frac{m}{y}+\frac{m}{xy}\\ &=1+\frac{y+mx+m}{xy}\\ &=1+\frac{1+m}{x(1-mx)}\\ &\ge1+\frac{1+m}{1/(4m)}~~~\color{gray}{\because x(1-mx)~\le~ 1/(4m) }\\ &=1+4m+4m^2\\ &=(2m+1)^2 \end{align}\] That means the minimum value of \(\left(1+\frac{1}{x}\right)\left(1+\frac{m}{y}\right)\) is \((2m+1)^2\) So, \((2m+1)^2= 81 \implies m =4,-5 \). Since they want positive value, discard \(-5\).
But \(m=4\) would certainly be incorrect since \(mx+y=1\) and \(x,y>0\).
try x = 1/8 y = 1/2
http://www.wolframalpha.com/input/?i=minimize++%5Cleft%28+1+%2B+%5Cfrac%7B1%7D%7Bx%7D+%5Cright%29%5Cleft%28+1+%2B+%5Cfrac%7B4%7D%7By%7D+%5Cright%29%2C+4x%2By%3D1%2Cx%3E0%2Cy%3E0
Oh I see!
Why \(\dfrac{y+mx+m}{xy}=\dfrac{1+m}{x(1-mx)}\) and why \(x(1-mx)\leq 4m\)?
I did my a super super long way... I differentiated: \[f(x)=(1+\frac{1}{x})(1+\frac{m}{1-mx}) \\ \text{ and then set } f'(x) =0 \\ \text{ which eventually led me to the solution } x=\frac{1}{2m} \\ \text{ then I found } m \text{ such that } f(\frac{1}{2m})=81 \]
for first question, just replace \(y+mx\) by \(1\) for second question, since \(m\) is positive, \(x(1-mx)\) is a parabola facing down with vertex at \((\frac{1}{2m}, \frac{1}{4m})\)

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