B = x2 - 2ax + a2 + x2 - 2bx + b2 + x2 - 2cx + c2
= 3x2 - 2(a + b + c)x + a2 + b2 + c2
= 3\(\left(x-\frac{a+b+c}{3}\right)^2\)- \(\frac{\left(a+b+c\right)^2}{3}\) + a2 + b2 + c2
B đạt min khi x = \(\frac{a+b+c}{3}\)
Thay x = \(\frac{a+b+c}{3}\)vào B
MinB = \(\left(\frac{a+b+c-3a}{3}\right)^2\)+ \(\left(\frac{a+b+c-3b}{3}\right)^2\) + \(\left(\frac{a+b+c-3c}{3}\right)^2\)
= \(\left(\frac{b+c-2a}{3}\right)^2\)+ \(\left(\frac{a+c-2b}{3}\right)^2\) + \(\left(\frac{a+b-2c}{3}\right)^2\)
Chỗ này mình làm hơi rối
B = \(3\left(x-\frac{a+b+c}{3}\right)^2-\frac{\left(a+b+c\right)^2}{3}\)+ a2 + b2 + c2
B đạt min khi x = (a + b + c)/3
MinB = a2 + b2 + c2 - \(\frac{\left(a+b+c\right)^2}{3}\)
