#### Question

If a,b,c are real numbers such that a+b+c=2, \dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}=\dfrac{1}{2}, show that one of numbers among a,b,c must be equal to 2

Collected in the board: Vieta's Formula

Steven Zheng posted 1 hour ago

Reduce to the common denominator for the given condition

\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}=\dfrac{ab+bc+ac}{abc} =\dfrac{1}{2}

Let m = ab+bc+ac, then abc = 2m

Then we have the following systems of equations

\begin{cases} a+b+c=2 \\ ab+bc+ac=m \\ abc = 2m \end{cases}

if and only if a,b.c are the three roots of the cubic equation x^3-2x+mx-2m = 0 by Vieta's formula.

Factorizing the left hand side of the equation gives

(x-2)(x^2+m)=0

Therefore, x=2 is always the root of the equation regardless of the value of m. So one of numbers among a,b,c must be equal to 2

Steven Zheng posted 1 hour ago

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