a, b, c are real numbers. such that a^2+b^2+c^2 = 1. prove that 1/2≤(ab+bc+ca)≤1

Question

its a algebra problem. please help

Answer 1

a^2 + b^2 + c^2 = 1

a, b, and c each have to be <=1 because if either was greater than 1, then a^2 + b^2 + c^2 > 1

a, b, and c each have to be >= -1 because if either was less than -1, then a^2 + b^2 + c^2 > 1

What we have now is:  -1 <= a, b, c <= 1

ab, bc, ca each have to be < 1 because if either was = 1, then both parts (a and b, b and c, or c and a) would have to be 1 and 1 or -1 and -1, which would make the squared parts (a^2 + b^2, b^2 + c^2, or c^2 + a^2) > 1

So the <=1 part of 1/2 <= ab + bc + ca <= 1 is incorrect.

Now consider the case of a=1, b=0, c=0.  That satisfies a^2 + b^2 + c^2 = 1.  But does it satisfy 1/2 <= ab + bc + ca ?

ab + bc + ca

1(0) + 0(0) + 0(0 = 0

0 is not greater than or equal to 1/2.

So the 1/2 <= part of 1/2 <= ab + bc + ca <= 1 is incorrect.

Answer:  a^2 + b^2 + c^2 = 1 does not mean 1/2 <= ab + bc + ca <= 1

Answer 2

a²+b²+c²=1 ‥Eq.1

a²+b²+c²-(ab+bc+ca)=½{(a²-2ab+b²)+(b²-2bc+c²)+(c²-2ca+a²)}=½{(a-b)²+(b-c)²+(c-a)²}

Here, (a-b)²≧0, (b-c)²≧0, (c-a)²≧0   So that, a²+b²+c²-(ab+bc+ca)≧0   Thus, we have:

ab+bc+ca≦a²+b²+c²   From Eq.1, we have: ab+bc+ca≦1 ‥Eq.2

While, a²+b²+c²+2(ab+bc+ca)=(a+b+c)²  

Here, (a+b+c)²≧0   So that, a²+b²+c²+2(ab+bc+ca)≧0   Thus, we have:

ab+bc+ca≧-½(a²+b²+c²)   From Eq.1, we have: ab+bc+ca≧-½ ‥Eq.3

Therefore, from Eq.2 and Eq.3, we have:

-½≦ab+bc+ca≦1   Q.E.D.