Finding minimum value of $xy$ given that $1/x + 1/y =2$?

Question

If for $x>0$ and $y>0$ we have $1/x + 1/y =2$. What can be the minimum value of $xy$? How do we do this simple question?

Answer 1

Use AM GM inequality

$$\dfrac{\dfrac1x+\dfrac1y}2\ge\sqrt{\dfrac1{xy}}$$ as $x,y>0$

the equality occurs if $\dfrac1x=\dfrac1y\implies x=y$

Alternatively,

for $x,y>0$

$$0\le\left(\dfrac1{\sqrt x}-\dfrac1{\sqrt x}\right)^2=\dfrac1x+\dfrac1y-\dfrac2{\sqrt{xy}}$$

$$\implies\dfrac2{\sqrt{xy}}\le\dfrac1x+\dfrac1y=2$$

$$xy\ge?$$

Answer 2

Of course, you can always try to use standard calculus techniques, like minimizing the function $P(x,y)=xy$ on the set $$ \{(x,y) \mid x>0,\ y>0,\ x^{-1}+y^{-1}=2 \}. $$ You do not really need Lagrange multipliers, since you can solve for $y$ and plug into $P$, getting a function of one real variable.

Answer 3

$1/y+1/x=2$.

$1/x^2+1/(xy)=2/x$.

$-1/x^2+2/x=1/xy.$

Completing the square:

$-(1/x -1)^2+1=1/(xy).$

$\Rightarrow 1/(xy) \le 1.$

$1 \le (xy).$

Minimum $xy=1$ for $1/x-1=0$.