If a and b are positive numbers, prove that the inequality square root of ab is less than or equal to 1/2 (a +

Question

If a and b are positive numbers, prove that the inequality square root of ab is less than or equal to 1/2 (a + b) as Euclid did, by considering a right triangle inscribed in a semicircle.

Answer 1

To prove this, I require a separate lemma (i.e. miniproof).

Claim: For positive numbers x and y, if x < y, then sqrt(x) < sqrt(y).

Proof: Let f(z) = sqrt(z). Then
f'(z) = 1/[2sqrt(z)]. If we set this to 0,
0 = 1/[2sqrt(z)]

Our critical numbers are ones which make f'(z) undefined, which would be z = 0. Should we test a value on the interval (0,infinity), we get a positive result. Therefore, the square root function is increasing on (0,infinity).

**end of Lemma**

Let a and b be positive numbers. Note that any number squared is greater than or equal to 0, i.e for any real number z,

z^2 >= 0

So let's substitute z for some other real value; let's say a - b

Therefore,

(a - b)^2 >= 0, so

a^2 - 2ab + b^2 >= 0

Adding 4ab to both sides, we get

a^2 + 2ab + b^2 >= 4ab

And now we can factor the left hand side as a perfect square.

(a + b)^2 >= 4ab

Since a and b are positive, then both sides of the inequality are positive. Therefore, if we take the square root, by the Lemma proven above,

sqrt [ (a + b)^2 ] >= sqrt (4ab)

On the left hand side, the square root of a square number is itself, eliminating the power of 2.

(a + b) >= sqrt(4ab)

We can pull the 4 out of the square root, to get

(a + b) >= 2sqrt(ab)

And if we multiply both sides by 1/2, we cancel the two on the right hand side, giving us

(1/2) [a + b] >= sqrt(ab)

OR

sqrt(ab) <= (1/2) [a + b]

Answer 2

Consider a semicircle with diameter AB and a point on the circumference C

From C drop a perpendicular to the circumference to meet it at D and let AD = a and BD = b

Now consider Δ's ACD and CBD

Since But
Therefore Also < CDA =
Therefore Δ's ACD and CBD are similar (equiangular)

Therefore, since corresponding sides are in proportion

So AD/CD = CD/BD

Thus CD² = AD.BD = ab

Thus CD = √(ab)

Now CD ≤ radius of the semicircle

AB = diameter = AD + DB = a + b

So radius = ½(a + b)

Therefore √(ab) ≤ ½(a + b)

And of course this says that the geometric mean of two numbers is less than or equal to their arithmetic mean. The equality holds when the two numbers are equal. (ie when D coincides with the centre of the circle)

Euclid was so elegant with his geometric proofs!!!!!!

Answer 3

let a and b be non-unfavorable actual numbers. Then sqrt(a) and sqrt(b) exist. when you consider that any decision squared is non-unfavorable we see that (sqrt(a) - sqrt(b))^2 => 0 which ability a - 2sqrt(ab) + b => 0 i.e. a + b => 2sqrt(ab) i.e. 2sqrt(ab) <= a + b i.e. sqrt(ab) <= (a+b)/2 as had to exhibit.