Given that u > 0 and v > 0 what is the smallest possible value of 1/u + 1/v given that u + v = 5?

Here is a solution from Danny of Milliken Mills High School, Canada.

Let S be the mininum value,
S = 1/u + 1/v = (v + u)/(uv).
Since u + v = 5 then S = 5/uv.

The maxinum value of uv gives a mininum value of S. Given u + v = 5 then v = 5 - u.

Let f(u) give the value of uv as u and v change. Then f(u) = u(5 - u) = -u ² + 5u . This is a quadratic function and the vertex (u, M) of the graph of f(u) is at (2.5, 6.25). It means the maxinum value occurs at M = 6.25 when u = 2.5 and v = 5 - u = 2.5. So the mininum value S = 1/2.5 + 1/2.5 = 0.8.

Vassil from Lawnswood High School, Leeds used the same method, and drew two graphs to illustrate it, one to show the possible values of u and v where u + v =5 and the other, shown below, the graph of f(u) = u(5 - u) = -u ² + 5u for the corresponding range of values of u and v.

Peter and Koopa of Boston College used the Arithmetic Mean - Geometric Mean inequality (AM-GM) noting as above that minimising 1/u + 1/v is the same as maximising uv. Here is Peter's solution:

From the AM-GM inequality:

The maximum of uv is therefore at equality, where u = v = 5/2, yielding max(uv) = 25/4

Hence the minimum value of 1/u + 1/v subject to u+v = 5, u, v > 0, is 4/5.