The known function f (x) satisfies: for any real number x1, X2, when Xi

The known function f (x) satisfies: for any real number x1, X2, when Xi

The most typical is f (x) = 2 ^ X
As long as it is an increasing function on R and f (x1 + x2) = f (x1) · f (x2) is OK

It is known that x1, X2, X3,..., x2010, x2011 belong to R
It is known that x1, X2, X3,..., x2010, x2011 belong to R. verification: X1 ^ 2 / x2 + x2 ^ 2 / X3 + X3 ^ 2 / X4 +... + x2010 ^ 2 / x1 ≥ X1 + x2 + X3 +... + x2010 + x2011
Are positive real numbers, there are wood, there are people, there are ideas!

It's very regular
Basic inequality
A + b > = 2 radical (a * b)
X1 ^ 2 / x2 + x2 > = 2 radical (x1 ^ 2 / x2 * x2) = 2 radical (x1 ^ 2) = 2x1 (x1 > 0)
When the equal sign holds, X1 ^ 2 / x2 = X2, that is, X1 = x2
In the same way
x2^2/x3+x3>=2x2
...
x2010^2/x2011 +x2011>=2 x2010
x2011^2/x1+x1>=2 x2011
Add left and right
have to
x1^2/x2+x2^2/x3+x3^2/x4+...+x2011^2/x1 + (x1+x2+...+x2011)>=2(x1+x2+...+x2011)
therefore
x1^2/x2+x2^2/x3+x3^2/x4+...+x2011^2/x1≥x1+x2+x3+...+x2010+x2011
The equal sign of the next certificate may hold
Because the first inequality is equal if X1 = X2, the second is x2 = X3,... And the last is x2011 = x1
So the condition of equal sign is X1 = x2 =... = x2011
therefore
x1^2/x2+x2^2/x3+x3^2/x4+...+x2011^2/x1≥x1+x2+x3+...+x2010+x2011

If x1, X2, X3 If the variance of x2010 and x2011 is 3, then 3 (x1-2), 3 (x2-2) 3 (x2010-2), 3 (x2011-2)
If x1, X2, X3 Then the variance is (x2-3)... (x2-010) What is the variance of 3 (x2010-2) and 3 (x2011-2)

Answer: 27