if a, b, c are three real numbers such that a^2+b^2+c^2=1, then ab+bc+ca lies in the interval__.?

Question

answer given is [-1/2, 1]. plz explain

Answer 1

ab + bc + ca =
(a^2 + b^2 + c^2) - (1/2)(a - b)^2 - (1/2)(b - c)^2 - (1/2)(c - a)^2
= 1 - (1/2)(a - b)^2 - (1/2)(b - c)^2 - (1/2)(c - a)^2
This shows that the maximum value of ab + bc + ca is 1 when a = b = c.

ab + bc + ca = (1/2)(a + b+ c)^2 - (1/2)(a^2 + b^2 + c^2)
= (1/2)(a + b+ c)^2 - (1/2)
This shows that the minimum value of ab + bc + ca is -1/2 when a + b + c = 0.

Answer 2

We have ab + bc + ac = ((a + b + c)^2 - (a^2 + b^2 + c^2))/2 = ((a + b + c)^2 - 1)/2. Therefore, ab + bc + ca is minimum when (a + b + c)^2 = 0, if this is possible. This happens if and only if a + b + c +0 and actually, if we set a =1,sqrt(2), b = -1/sqrt(2) and c=0, we satisfy a^2+b^2+c^2=1and get a + b + c =0. Therefore, the minimum value ab + bc + ac can take is |(0^2 -1)/2 = -1/2.

To find the maximum, we have to find the maximum value of (a + b + c)^2 subject to a^2+b^2+c^2=1. Since (a + b + c)^2 = |a + b + c|^2 and |a + b + c | <= |a| + |b| + |c|, it's enough to consider positive values for a, b and c. Therefore, it's enough to maximize a + b + c

Using Lagrange multipliers, define g(a,b,c , L) = a + b + c - L(a^2 + b^2 + c^2). Differentiating with respect to a, b , c and L setting these partial derivatives to 0, , we get

1 - 2aL = 0 => L = 1/2a
1 - 2bL = 0 => L = 1/2b
1 - 2cL = 0 => L = 1/2c
a^2 + b^2 + c^2 =1

Therefore, we have a = b = c = 1/sqrt(3) and ab + bc + ac = 3/3 = 1 . This is the only positive solution. In virtue of the differentiability of the functions involved, if there were another extremum point there would be another solution for this system. Since this is not a minimum point, it's a relative maximum. But on the boundary of the feasible set, that is, when when of the variables is 1 or -1, we get ab + bc + ac = 0 <1. Therefore, the maximum values is 1.

And ab + bc + ac lies in [-1/2, 1]. This is an analytic proof.

Answer 3

ab + bc+ ca = half the surface area of a rectangular prism (box) a x b x c.
Where a^2 + b^2 + c^2 is the diagonal^2 of the same rectangular prism.
A cube would have the lowest diagonal /surface area ratio.
where a=b=c = 1/sqrt3.,, ab+bc+ca = 1
A line where a and b are zero would have the highest diagonal to surface area ratio = 1/0.,, ab+bc+ca = 0
I can't account for the negative surface area.

Answer 4

enable xa = (a+b+c) = 0 --- --- --- --- --- --- --- --- (a million); ya = xa^2 = (a+b+c)^2 = a^2 +b^2 + c^2 +2ab + 2bc + 2ca --- --- --- --- --- --- --- --- (2); ya = 0; So, from (2), a^2 +b^2 + c^2 = -2(ab + bc + ca) --- --- --- --- --- --- --- --- (3); enable, p = a^2 +b^2 + c^2, --- --- --- --- --- --- --- --- (4); and q = -(ab + bc + ca) --- --- --- --- --- --- --- --- (5); From (3), p/q = 2 --- --- --- --- --- --- --- --- --- --- --- --- --- (6); Now, enable us to dissect the LHS with its words; From (a million), (a+b+c) = 0 --- --- --- --- --- --- --- --- (a million); Multiply eqn (a million) by technique of a; We get, a(a+b+c) = 0 that's, a^2 + ab + ca = 0 that's, a^2 - bc = -(ab + bc+ ca) --- --- --- --- --- --- --- --- (7); further, multiplying eqn (a million) by technique of b, we are able to get, b^2 - ca = -(ab + bc+ ca) --- --- --- --- --- --- --- --- (8); And back, multiplying eqn (a million) by technique of c, we are able to get, c^2 - ab = -(ab + bc+ ca) --- --- --- --- --- --- --- --- (9); as a result, making use of (7), (8) & (9), we are able to rearrange the LHS as follows; LHS = a^2/(a^2-bc) + b^2/(b^2-ca) + c^2/(c^2-ab) = (a^2 +b^2+ c^2)/(-ab-bc-ca) --- --- --- --- (10); From (4), (5) & (10), we get, LHS = p/q --- --- --- --- --- --- --- --- (11); And, from (6) & (11), we get, LHS = 2 that's, a^2/(a^2-bc) + b^2/(b^2-ca) + c^2/(c^2-ab) = 2 --- --- --- --- --- --- --- --- (proved);