Question

Given a,b,c are positive real numbers such that

a+b+c = \sqrt{10+\sqrt{19} }

and

\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c} = \sqrt{10-\sqrt{19} }

If x = a^2+b^2+c^2, find the value of x+\dfrac{9}{x}

Collected in the board: Inequality

Steven Zheng posted 10 hours ago

Answer

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

=3+\dfrac{a}{b}+\dfrac{b}{a}+\dfrac{b}{c}+\dfrac{c}{b}+\dfrac{a}{c}+\dfrac{c}{a}

=9+\bigg( \sqrt{\dfrac{a}{b}}-\sqrt{\dfrac{b}{a} } \bigg) ^2+ \bigg( \sqrt{\dfrac{b}{c}}-\sqrt{\dfrac{c}{b} }\bigg) ^2+\bigg( \sqrt{\dfrac{a}{c}}-\sqrt{\dfrac{c}{a} } \bigg) ^2

\geq 9

which shows (a+b+c)(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c} ) = 9, if and only if a=b=c

On the other hand,

Using given conditions

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

= \sqrt{10+\sqrt{19} }\cdot \sqrt{10-\sqrt{19} }

=\sqrt{10^2-19}

=9

Therefore

a=b=c

Then

3a = \sqrt{10+\sqrt{19} }

a^2 = \dfrac{10+\sqrt{19} }{9}

Similarly,

\dfrac{3}{a} = \sqrt{10-\sqrt{19} }

\dfrac{1}{a^2} = \dfrac{10-\sqrt{19} }{9}

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

x = a^2+b^2+c^2 = 3a^2

\dfrac{1}{x} = \dfrac{1}{3a^2}

Therefore,

x+\dfrac{9}{x} = 3a^2+ \dfrac{9}{3a^2}

=\dfrac{10+\sqrt{19} }{3}+\dfrac{9}{3}\cfrac{}{} \dfrac{10-\sqrt{19} }{9}

=\dfrac{20}{3}

Steven Zheng posted 10 hours ago

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