# Question #90052

Dec 10, 2016

${x}^{n} + \frac{1}{a} + {y}^{n} + \frac{1}{b} = 2 {\left(\frac{a b}{a + b}\right)}^{n} + \frac{1}{a} + \frac{1}{b}$

#### Explanation:

Assuming that the correct formulation is

If $\frac{x}{a} + \frac{y}{b} = 1 \mathmr{and} {x}^{2} / a + {y}^{2} / b = \frac{a b}{a + b}$

then ${x}^{n} + \frac{1}{a} + {y}^{n} + \frac{1}{b} =$

From

$\left\{\begin{matrix}\frac{x}{a} + \frac{y}{b} = 1 \\ {x}^{2} / a + {y}^{2} / b = \frac{a b}{a + b}\end{matrix}\right.$

solving for $x , y$ we obtain

$x = y = \frac{a b}{a + b}$

To obtain this result we solve for $X = \frac{x}{a} , Y = \frac{y}{b}$

$\left\{\begin{matrix}X + Y = 1 \\ a {X}^{2} + b {Y}^{2} = \frac{a b}{a + b}\end{matrix}\right.$

$a {X}^{2} + b {\left(1 - X\right)}^{2} = \frac{a b}{a + b}$ and solving for $X$

$\left(a + b\right) {X}^{2} - 2 b X + b = \frac{a b}{a + b}$ or

${X}^{2} - \frac{2 b}{a + b} X + {\left(\frac{b}{a + b}\right)}^{2}$ or

${\left(X - \frac{b}{a + b}\right)}^{2} = 0$

then $X = \frac{b}{a + b}$ and analogously $Y = \frac{a}{a + b}$

so $x = y = \frac{a b}{a + b}$

Finally we have

${x}^{n} + \frac{1}{a} + {y}^{n} + \frac{1}{b} = 2 {\left(\frac{a b}{a + b}\right)}^{n} + \frac{1}{a} + \frac{1}{b}$