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Question Number 5018 by lec123 last updated on 02/Apr/16

If (a−b)^2 +(b−c)^2 +(c−a)^2 =0, then  the values of a, b and c are equal.

$$\mathrm{If}\:\left({a}−{b}\right)^{\mathrm{2}} +\left({b}−{c}\right)^{\mathrm{2}} +\left({c}−{a}\right)^{\mathrm{2}} =\mathrm{0},\:\mathrm{then} \\ $$$$\mathrm{the}\:\mathrm{values}\:\mathrm{of}\:{a},\:{b}\:\mathrm{and}\:{c}\:\mathrm{are}\:\mathrm{equal}. \\ $$

Commented by Rasheed Soomro last updated on 03/Apr/16

a=b=c only when a,b,c ∈ R.

$${a}={b}={c}\:{only}\:{when}\:{a},{b},{c}\:\in\:\mathbb{R}. \\ $$

Answered by prakash jain last updated on 02/Apr/16

If (a−b)^2  is a whole square so (a−b)^2 ≥0.  Similarly (b−c)^2 ≥0 and (c−a)^2 ≥0  Given (a−b)^2 +(b−c)^2 +(c−a)^2 ≥0  Each of the terms have to be equal to 0.  a−b=0⇒a=b  b−c=0⇒b=c  c−a=0⇒c=a  Hence  a=b=c

$$\mathrm{If}\:\left({a}−{b}\right)^{\mathrm{2}} \:\mathrm{is}\:\mathrm{a}\:\mathrm{whole}\:\mathrm{square}\:\mathrm{so}\:\left({a}−{b}\right)^{\mathrm{2}} \geqslant\mathrm{0}. \\ $$$$\mathrm{Similarly}\:\left({b}−{c}\right)^{\mathrm{2}} \geqslant\mathrm{0}\:\mathrm{and}\:\left({c}−{a}\right)^{\mathrm{2}} \geqslant\mathrm{0} \\ $$$$\mathrm{Given}\:\left({a}−{b}\right)^{\mathrm{2}} +\left({b}−{c}\right)^{\mathrm{2}} +\left({c}−{a}\right)^{\mathrm{2}} \geqslant\mathrm{0} \\ $$$$\mathrm{Each}\:\mathrm{of}\:\mathrm{the}\:\mathrm{terms}\:\mathrm{have}\:\mathrm{to}\:\mathrm{be}\:\mathrm{equal}\:\mathrm{to}\:\mathrm{0}. \\ $$$${a}−{b}=\mathrm{0}\Rightarrow{a}={b} \\ $$$${b}−{c}=\mathrm{0}\Rightarrow{b}={c} \\ $$$${c}−{a}=\mathrm{0}\Rightarrow{c}={a} \\ $$$$\mathrm{Hence} \\ $$$${a}={b}={c} \\ $$

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