Asked by Mirie
Prove: [1/sqrt(2)] [sqrt(a) + sqrt(b)] <= sqrt(a + b) <= sqrt(a) + sqrt(b) for all non-negative real numbers a and b.
Answers
Answered by
Steve
1/√2 (√a + √b) <= √(a+b)
1/2 (a + 2√ab + b) <= a+b
√ab <= (a+b)/2
ab <= a^2/2 + ab + b^2/2
clearly true
√(a+b) <= √a + √b
a+b <= a + 2√ab + b
clearly true
1/2 (a + 2√ab + b) <= a+b
√ab <= (a+b)/2
ab <= a^2/2 + ab + b^2/2
clearly true
√(a+b) <= √a + √b
a+b <= a + 2√ab + b
clearly true
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