Given x＞0，y＞0, 1/x ＋ 8/$y^2$ ＝1, solve the minima of $x$＋$y$?
One guy has this problem solved by Cauchy inequality.
Can anyone have another approach or?
I worked out like this: if 1/$x$ ＝ 8/$y^2$ ＝ 1/2 hold, then $x$＝2, $y$＝4， (1/$x$)(8/$y^2$)＝1/4 ⥤ $x$$y^2$＝32 ⥤$ xy$＝8⥤ $x$＋$y$＝2+4＝6
Is there anything wrong with my approach? If yes, please advice and explain.
Thanks
Let $s:=x+y$ and let us eliminate $y$.
$$\frac1x+\frac8{y^2}=1$$ becomes
$$\frac1x+\frac8{(s-x)^2}=1$$ or
$$s=x+\sqrt{\frac{8x}{x-1}}.$$
The derivative cancels for $x=2$, and $s=6$.