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Numerical Ability
If a2 + b2 + c2 = 1, then ab + bc + ac lies in the interval
Read Solution (Total 1)
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- using: (a + b + c)^2 = a^2 + b^2 + c^2 + 2(ab + bc + ca)
(a + b + c)^2 ≥ 0 , then:
a^2 + b^2 + c^2 + 2(ab + bc + ca) ≥ 0
1 + 2(ab + bc + ca) ≥ 0
2(ab + bc + ca) ≥ -1
ab + bc + ca ≥ -1/2
now using: (a - b)^2 + (b - c)^2 + (c - a)^2 = 2(a^2 + b^2 + c^2) - 2(ab + bc + ca)
(a - b)^2 + (b - c)^2 + (c - a)^2 ≥ 0
2(a^2 + b^2 + c^2) - 2(ab + bc + ca) ≥ 0
2(1) - 2(ab + bc + ca) ≥ 0
-2(ab + bc + ca) ≥ -2
ab + bc + ca ≤ 1
hence:
-1/2 ≤ ab + bc + ca ≤ 1
[-1/2 , 1] - 9 years agoHelpfull: Yes(0) No(0)
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