Find the minimum value of $(x+y)$

Two positive numbers $x$ and $y$ vary in such a way that

$\ 128x^2-16x^2y+1=0$

Find the minimum value of $(x+y)$.

• Lagrange Multipliers is one way. Still, are you sure the $1$ is not a $-1$? – Gary. Aug 22 '15 at 16:00
• Yea, I checked the question paper again. Also, I'm only in a Secondary School (in Singapore) so I haven't learnt all these methods yet. I've only learnt a fair bit of calculus so that's all I can use to solve the question. – Sheow Boon Aug 22 '15 at 16:12

$$y=8+\dfrac1{16x^2}$$
$$\implies x+y=x+8+\dfrac1{16x^2}$$
As $x>0,$ using AM, GM inequality $$x+\dfrac1{16x^2}=\dfrac x2+\dfrac x2+\dfrac1{16x^2}=\dfrac{\dfrac x2+\dfrac x2+\dfrac1{16x^2}}3\ge\sqrt[3]{\dfrac x2\cdot\dfrac x2\cdot\dfrac1{16x^2}}=\dfrac14$$
• @SheowBoon, Use en.wikipedia.org/wiki/Second_derivative_test on $$f(x)=x+8+\dfrac1{16x^2}$$ – lab bhattacharjee Aug 22 '15 at 16:22