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