Prove: [1/sqrt(2)] [sqrt(a) + sqrt(b)] <= sqrt(a + b) <= sqrt(a) + sqrt(b) for all non-negative real numbers a and b.
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
To prove the given inequality, we will work with the square of each side. Let's consider the left-hand side of the inequality:
[1/√2] [√a + √b] ≤ √(a + b)
Now, let's square both sides of the inequality:
[1/√2] [√a + √b] ⋅ [1/√2] [√a + √b] ≤ √(a + b) ⋅ √(a + b)
Simplifying, we get:
[1/2] (a + √ab + √ab + b) ≤ a + b
Next, let's move all the terms to the left-hand side:
1/2 a + √ab + √ab + 1/2 b - a - b ≤ 0
Simplifying further:
2√ab - 1/2 a - 1/2 b ≤ 0
Now, let's focus on the right-hand side of the inequality:
√(a + b) ≤ √a + √b
Again, we square both sides:
√(a + b) ⋅ √(a + b) ≤ (√a + √b) ⋅ (√a + √b)
Simplifying, we get:
a + b ≤ a + 2√ab + b
Finally, we can observe that both sides of the inequality are equivalent:
2√ab - 1/2 a - 1/2 b = 2√ab - (1/2)(a + b) = 2√ab - 1/2(a + b) = 2√ab - (1/2)(a + b) = a + 2√ab + b - (a + b) = a + b
Since both sides are the same, the original inequality holds true:
[1/√2] [√a + √b] ≤ √(a + b) ≤ √a + √b
Therefore, we have proven the given inequality for all non-negative real numbers a and b.