﻿ Given a,b are positive real numbers such that a+b= \sqrt{4+2\sqrt{3} } and \dfrac{1}{a}+\dfrac{1}{b} = \sqrt{4-2\sqrt{3} }

#### Question

Given a,b are positive real numbers such that

a+b= \sqrt{4+2\sqrt{3} }

and

\dfrac{1}{a}+\dfrac{1}{b} = \sqrt{4-2\sqrt{3} }

If x = a^2+b^2, find the value of x+\dfrac{4}{x}

Collected in the board: Inequality

Steven Zheng posted 4 months ago

#### Answer

(a+b)(\dfrac{1}{a}+\dfrac{1}{b} )

=2+\dfrac{a}{b}+\dfrac{b}{a}

=2+\bigg( \sqrt{\dfrac{a}{b}}-\sqrt{\dfrac{b}{a} } \bigg) ^2

\geq 4

which shows (a+b)(\dfrac{1}{a}+\dfrac{1}{b} ) = 4, if and only if a=b=c

On the other hand,

Using given conditions

(a+b)(\dfrac{1}{a}+\dfrac{1}{b} )

= \sqrt{4+2\sqrt{3} }\cdot \sqrt{4-2\sqrt{3} }

=\sqrt{4^2-12}

=4

Therefore

a=b

Then

2a = \sqrt{4+2\sqrt{3} }

a^2 = \dfrac{4+2\sqrt{3} }{4}

Similarly,

\dfrac{2}{a} = \sqrt{4-2\sqrt{3} }

\dfrac{1}{a^2} = \dfrac{4-\sqrt{3} }{4}

Represent x and \dfrac{1}{x} in terms of a

x = a^2+b^2= 2a^2

\dfrac{1}{x} = \dfrac{1}{2a^2}

Therefore,

x+\dfrac{4}{x} = 2a^2+ \dfrac{4}{2a^2}

=\dfrac{4+2\sqrt{3} }{2}+\dfrac{4}{2}\cdot \dfrac{4-2\sqrt{3} }{4}

=\dfrac{8}{2} = 4

Steven Zheng posted 4 months ago

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