Given p and q are positve real numbers and 1/p +1/q = 1..

[rsa58]

hey i can't figure this out:

if p and q are positve real numbers and

1/p +1/q = 1

show that if u and v are greater than or equal to zero then

uv=< (u^p)/p +(v^q)/q.

 

[morphism]

http://en.wikipedia.org/wiki/Young's_inequality

 

[tacman]

To show that uv is less than or equal to (u^p)/p + (v^q)/q, we will use the AM-GM inequality. The AM-GM inequality states that for any two positive real numbers a and b, the arithmetic mean (a+b)/2 is always greater than or equal to the geometric mean √(ab). This means that (a+b)/2 ≥ √(ab).

Using this inequality, we can rewrite the left side of the given equation as:

1/p + 1/q = (1/2)(1/p + 1/q) + (1/2)(1/p + 1/q)

= (1/2)(1/p + 1/p) + (1/2)(1/q + 1/q)

= (2/p) + (2/q)

= 2(1/p + 1/q)

= 2(1)

= 2

Now, let's look at the right side of the given equation:

(u^p)/p + (v^q)/q = (1/p)(u^p) + (1/q)(v^q)

Using the AM-GM inequality, we can rewrite this as:

(1/p)(u^p) + (1/q)(v^q) ≥ 2√((u^p)(v^q))

= 2√((uv)^p)

= 2(uv)^p/q

= 2(uv)^1

= 2uv

Since 2 ≥ 2uv, we can conclude that:

1/p + 1/q ≥ (u^p)/p + (v^q)/q

which means that:

uv ≤ (u^p)/p + (v^q)/q

This shows that for any two positive real numbers u and v, if 1/p + 1/q = 1, then uv is less than or equal to (u^p)/p + (v^q)/q. Therefore, the given statement is true.