Let a = (x1 + x2 +... X2010) * X1 + x2 +... X2010), B = (x1 + x2 +... X2010) * (x1 + x2 +... X2010) Let a = (x1 + x2 +... X2010) * x2 + X3 +... X2011), B = (x1 + x2 +... X2011) * (x2 + X3 +... X2011) * (x1 + X3 +... X2011) * (x1 + x2... + x2010), and compare the size of a and B

Let a = (x1 + x2 +... X2010) * X1 + x2 +... X2010), B = (x1 + x2 +... X2010) * (x1 + x2 +... X2010) Let a = (x1 + x2 +... X2010) * x2 + X3 +... X2011), B = (x1 + x2 +... X2011) * (x2 + X3 +... X2011) * (x1 + X3 +... X2011) * (x1 + x2... + x2010), and compare the size of a and B

a＞b M=x2+x3+… +Then a = (x1 + m) (M + x2011), B = (x1 + m + x2011) * M

Given that {x1, X2, X3, X4} is contained in {x > 0 (x-3) sin π x = 0}, then the minimum value of X1 + x2 + X3 + X4 is
What if it's equal to one

(x-3)sinπx=0,
We obtain x-3 = 0 or sin π x = 0
The solution is x = 3, or π x = k, π, K ∈ n +,
That is, x = 1,2,3,4
So {x > 0 (x-3) sin π x = 0} = n +,
So the minimum value of X1 + x2 + X3 + X4 is 1 + 2 + 3 + 4 = 10

Given that f (x) = sin (x + π / 4), if the equation f (x) = m on [0,2 π] has two unequal real roots x1, X2, then the value of X1 + X2 is
How many?

Is it. 5 π or 2.5 π?
I use the translation of the image to make the image
Because there are two unequal real roots in the title, the range of X is (0,. 5 π) or (. 75 π, 7 / 4 π)
There are two solutions