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 $$ x+\dfrac{9}{x} $$

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 $$ x+\dfrac{9}{x}  $$

Uniteasy Admin · Accepted 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,
$$x+\dfrac{9}{x} = 3a^2+  \dfrac{9}{3a^2} $$
$$=\dfrac{10+\sqrt{19} }{3}+\dfrac{9}{3}\cfrac{}{} \dfrac{10-\sqrt{19} }{9} $$
$$=\dfrac{20}{3} $$

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