Prove the following inequalities. Then state for which values the LHS equals RHS.
1) a^2 + b^2 ≥ 2(a-b-1)
2) Given a + b = 1, prove the inequality a^2 + b^2 ≥ 1/2
Please explain how you solved it. Thanks.
To prove the given inequalities, let's expand the expressions on the left-hand side (LHS) and then rearrange the terms:
1) a^2 + b^2 ≥ 2(a - b - 1)
Expanding the LHS:
a^2 + b^2 ≥ 2a - 2b - 2
Rearranging terms:
a^2 - 2a + b^2 + 2b ≥ -2
Completing the squares for a and b:
(a^2 - 2a + 1) + (b^2 + 2b + 1) ≥ -2 + 1 + 1
(a - 1)^2 + (b + 1)^2 ≥ 0
Since (a - 1)^2 ≥ 0 and (b + 1)^2 ≥ 0 for any real values of a and b, the sum of these squares will always be greater than or equal to zero. Hence, the inequality holds for all values of a and b. There are no specific values for which the LHS equals the RHS.
2) Given a + b = 1, prove the inequality a^2 + b^2 ≥ 1/2
By substituting a + b = 1 into the inequality:
(a + b)^2 ≥ 1/2
Expanding the LHS:
a^2 + 2ab + b^2 ≥ 1/2
Since a + b = 1, substitute 1 for a + b in the expression:
a^2 + 2ab + b^2 ≥ 1/2
Rearranging terms:
a^2 + b^2 + 2ab ≥ 1/2
We can rewrite the left-hand side as (a + b)^2:
(a + b)^2 - 2ab ≥ 1/2
Substituting 1 for a + b:
(1)^2 - 2ab ≥ 1/2
Simplifying:
1 - 2ab ≥ 1/2
Multiplying through by 2:
2 - 4ab ≥ 1
Rearranging terms:
4ab ≤ 1
Dividing both sides by 4:
ab ≤ 1/4
Since the product of two positive real numbers is always non-negative, the inequality holds for all values of a and b that satisfy a + b = 1. In other words, for any values a and b that sum up to 1, the inequality a^2 + b^2 ≥ 1/2 holds true.