# Find the minimum value of $x+2y$ given $\frac{1}{x + 2} + \frac{1}{y + 2} = \frac{1}{3}.$

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!

• 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. – Ripi2 May 23 '20 at 23:47
• Should this have the algebra-precalculus tag? – Bladewood May 23 '20 at 23:51

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 $$3+6\sqrt{2}.$$
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.
• I got $y=\dfrac{x+8}{x-1}$ – J. W. Tanner May 24 '20 at 2:25