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Let $x$ and $y$ be positive real numbers such that $$\frac{1}{x + 2} + \frac{1}{y + 2} = \frac{1}{3}.$$Find the minimum value of $x + 2y.$


I think I will need to use the Cauchy-Schwarz Inequality here, but I don't know how I should use it. Can anyone help?

Thanks!

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  • $\begingroup$ Can you write the first equation as $y=f(x)$? Sure you can. Then substitute this new form of y into the second equation. Then derive and make equal to zero. Solve for x. With this x get y from the first equation. Operate the second and you're done. $\endgroup$ – Ripi2 May 23 '20 at 23:47
  • $\begingroup$ Should this have the algebra-precalculus tag? $\endgroup$ – Bladewood May 23 '20 at 23:51
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Cauchy-Schwarz implies $$((x+2)+2(y+2))\left(\frac 1{x+2}+\frac 1{y+2}\right)\geq (1+\sqrt{2})^2 $$ $$\Rightarrow x+2y+6\geq 3(1+\sqrt{2})^2,$$where equality is achieved when $$x+2=3(1+\sqrt{2}),y+2=\frac 3{\sqrt{2}}(1+\sqrt{2}).$$ This shows that the minimum of $x+2y$ is $3+6\sqrt{2}.$

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Hint: $y = \left(\dfrac{1}{3} - \dfrac{1}{x+2}\right)^{-1}-2= \dfrac{3x+6}{x-1}-2= \dfrac{x+8}{x-1}\implies x+2y=x+\dfrac{2x+16}{x-1}= \dfrac{x^2+x+16}{x-1}=f(x)$.From this point, you simply set $f'(x) = 0$ and solve for critical points and take it from there. It should be standard calculus problem.

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    $\begingroup$ I got $y=\dfrac{x+8}{x-1}$ $\endgroup$ – J. W. Tanner May 24 '20 at 2:25

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