關於數學問題 factorization

(a+b+c^2) = 3a^2 + 3b^2 + 3c^2

=> 3a^2 + 3b^2 + 2c^2 - a - b = 0

The above equation is in fact a ellipsoid with infinite number of intersection point with a+b+c = 12.

Your question is probably something like:

(a+b+c)^2 = 3(a^2+b^2+c^2).

Expand and eliminate equal term gives:

2(a^2+b^2+c^2) = 2ab+2bc+2ca
a^2 + b^2 + c^2 = ab + bc + ca

However, by rearrangement inequality a^2 + b^2 + c^2 >= ab + bc + ca
because without lossing generality we can assume a>=b>=c.
a^2 + b^2 + c^2 is the direct sum (max * max, middle * middle, min*min)
ab + bc + ca is the random sum, and direct sum >= random sum

Equality holds only iff a = b = c.
As a^2 + b^2 + c^2 = ab + bc + ca, we can conclude a = b = c.

As a + b + c = 12, a = 4.

(a+b+c^2) = 3a^2 + 3b^2 + 3c^2

=> 3a^2 + 3b^2 + 2c^2 - a - b = 0

The above equation is in fact a ellipsoid with infinite number of intersection point with a+b+c = 12.
Expand and eliminate equal term gives:

2(a^2+b^2+c^2) = 2ab+2bc+2ca
a^2 + b^2 + c^2 = ab + bc + ca

However, by rearrangement inequality a^2 + b^2 + c^2 >= ab + bc + ca
because without lossing generality we can assume a>=b>=c.
a^2 + b^2 + c^2 is the direct sum (max * max, middle * middle, min*min)
ab + bc + ca is the random sum, and direct sum >= random sum

Equality holds only ifa = b = c.
As a^2 + b^2 + c^2 = ab + bc + ca, we can conclude a = b = c.
so, a + b + c = 12, a = 4.

係 (a + b + c2) = 3(a2 + b2 + c2)
還是 (a + b + c)2 = 3(a2 + b2 + c2)？