Hi and welcome to the Matladpi blog!

Let there be two numbers a and b such that their sum is n, n > 0. What numbers should you pick in order to maximize the number a*b?

Answer:

a = a

a + b = n <=> b = n - a

=>

a*b = a(n - a) = na - a^2 = f(a)

f'(a) = 0

n - 2a = 0

2a = n

a = n/2

n/3 < n/2

f'(n/3) = n - (2n)/3 = n/3 > 0

2n/3 > n/2

f'(2n/3) = n - (4n)/3 = -n/3 < 0

=> f(n/2) is the maximum

b = n - a = n - n/2 = n/2

=> maximum number a*b = (n^2)/4

Let there be two positive numbers a and b. Show that their geometric mean is at most their arithmetic mean.

Answer:

(ab)^(1/2) <= (a + b)/2

ab <= (a^2 + 2ab + b^2)/4

a^2/4 + ab/2 - ab + b^2/4 >= 0

a^2/4 - ab/2 + b^2/4 >= 0

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

(a - b)^2 >= 0

True for all positive numbers a and b. The equality happens when a = b.

Thus,

(ab)^(1/2) <= (a + b)/2

I am Jesse Sakari Hyttinen and I will see you in the next post!

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