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Question Number 213391 by York12 last updated on 04/Nov/24
Solve the system of equations where a,b,c≥0  a−2bc=b−2ac=c−2ab  a+b+c=2
$$\mathrm{Solve}\:\mathrm{the}\:\mathrm{system}\:\mathrm{of}\:\mathrm{equations}\:\mathrm{where}\:{a},{b},{c}\geqslant\mathrm{0} \\ $$$${a}−\mathrm{2}{bc}={b}−\mathrm{2}{ac}={c}−\mathrm{2}{ab} \\ $$$${a}+{b}+{c}=\mathrm{2}\: \\ $$
Answered by Frix last updated on 04/Nov/24
Due to symmetry a=b=c=(2/3)
$$\mathrm{Due}\:\mathrm{to}\:\mathrm{symmetry}\:{a}={b}={c}=\frac{\mathrm{2}}{\mathrm{3}} \\ $$
Commented by York12 last updated on 04/Nov/24
yes you are right sorry, does lagrange  multipliers necessarily gives all equality cases
$$\mathrm{yes}\:\mathrm{you}\:\mathrm{are}\:\mathrm{right}\:\mathrm{sorry},\:\mathrm{does}\:\mathrm{lagrange} \\ $$$$\mathrm{multipliers}\:\mathrm{necessarily}\:\mathrm{gives}\:\mathrm{all}\:\mathrm{equality}\:\mathrm{cases} \\ $$
Commented by mr W last updated on 04/Nov/24
solutions:  ((2/3), (2/3), (2/3))  (3, −(1/2), −(1/2))  (6, −2, −2)
$${solutions}: \\ $$$$\left(\frac{\mathrm{2}}{\mathrm{3}},\:\frac{\mathrm{2}}{\mathrm{3}},\:\frac{\mathrm{2}}{\mathrm{3}}\right) \\ $$$$\left(\mathrm{3},\:−\frac{\mathrm{1}}{\mathrm{2}},\:−\frac{\mathrm{1}}{\mathrm{2}}\right) \\ $$$$\left(\mathrm{6},\:−\mathrm{2},\:−\mathrm{2}\right) \\ $$
Commented by Frix last updated on 04/Nov/24
Yes but a, b, c ≥0
$$\mathrm{Yes}\:\mathrm{but}\:{a},\:{b},\:{c}\:\geqslant\mathrm{0} \\ $$
Commented by mr W last updated on 04/Nov/24
i just read the truncated question :(
$${i}\:{just}\:{read}\:{the}\:{truncated}\:{question}\::\left(\right. \\ $$

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