Division by 0? Nah

Algebra Level 5

Positive real numbers $x$ and $y$ are such that $x>y$ and $xy=1$ . Find the minimum value of $A=\dfrac { { x }^{ 2 }+{ y }^{ 2 } }{ x-y }$ .

If the answer can be expressed in the form of $\sqrt [ a ]{ b }$ , where $b$ is a positive integer and $a$ is a positive real, find the minimum value of $A+a+b$ .

Write your answer to 3 decimal places.

The answer is 5.495.

1 solution

Chew-Seong Cheong
Jul 15, 2017

Relevant wiki: Arithmetic Mean - Geometric Mean

$\begin{aligned} A & = \frac {x^2+y^2}{x-y} \\ & = \frac {(x-y)^2+2xy}{x-y} \\ & = x-y + \frac 2{x-y} & \small \color{#3D99F6} \text{Since }x-y >0 \text{, we can use AM-GM inequality.} \\ & \ge 2\sqrt 2 = \sqrt [\frac 23] 2 \end{aligned}$

Equality occurs when $x-y = \dfrac 2{x-y}$ , or $x = \frac {1+\sqrt 3}{\sqrt 2}, y = \frac {\sqrt 2}{1+\sqrt 3}$ .

$\implies A + a + b = 2\sqrt 2 + \dfrac 23 + 2 \approx \boxed{5.495}$ .

That's a great solution!

Anyways, it's still better to show why $b=2$ and $a=2/3$ , which is easy as $b$ is an integer.

Steven Jim - 3 years, 11 months ago

Shouldn't 2 times root 2 be equal to 2^(3/2) ?? And hence b=2, a= 3/2 ??

Aaghaz Mahajan - 3 years, 10 months ago

$\sqrt[x]a = a^\frac 1x \implies a^x = \sqrt[\frac 1x]a$ .

Chew-Seong Cheong - 3 years, 10 months ago

Oh Thanks!! I did it correctly but kept on inputting the wrong value of a !!

Aaghaz Mahajan - 3 years, 10 months ago

Division by 0? Nah

The answer is 5.495.

1 solution

0 pending reports