Lower bound of estimated minimum value (that is, what is the minimum value of u at least):

First [a, b] < =ab, [b, c] <; =bc，[c，a]& lt； = About.

Let k = a+b+C b+c.

There are:

u & gt= k/2-(a b+ BC+ca)/k & gt； =k/2-(a^2+b^2+c^2)/k>； =k/2-k^3/(3k)=k/6.

Therefore, even if u reaches the minimum value, it cannot be less than k/6.

We certainly hope that u=k/6. Then when k takes the minimum value, u takes the minimum value.

First of all, a, b, c & gt=2.

So k>= 6.u>= 1.

At this time, as long as u is closest to 1 or directly close to 1, this u value is the minimum value.

When k=6, u takes 2. Therefore, the minimum value of u can only be between 1 and 2. As can be seen from the above inequality, when k >: At 12, u >；; 2. So the minimum value can only be 6.

If (a, b)=(b, c)=(c, a)= 1, then the minimum value of u is among the corresponding values of such an array (a, b, c). In this way, the following enumeration can quickly find out the minimum value of each k:

K=6 has a unique u=2.

K=7 has a unique u=3/2.

K=8 has the smallest u= 17/8.

K=9 has the smallest u=37/ 18.

K= 10 minimum u= 19/ 10.

K= 1 1 has the smallest u=53/22.

K= 12 has a minimum value of u=2.

The smallest one is the value when k=7 and U = 3/2. A = B = 2 and C = 3 are corresponding arrays, so the minimum value of u is 3/2.