Respuesta :
Answer: [tex]a+b+c=\frac{n+1}{n}.[/tex]
Step-by-step explanation: The function [tex]f(x_1,x_2,\ldots)=x_1^2+x_2^2+\ldots[/tex] is always positive except at the origin where it is equal to zero. This means that the absolute minumum of this function must be [tex]a=0[/tex]. Absolute maximum is when all of the variables are equal to zero except [tex]x_1[/tex] which is equal to 1 (f evaluated at this point is equal to 1 do b=1). The function itself is then equal to 1. This is because when [tex]f(\cdots)=x_1^2+x_2^2+\ldots\leq x_1^2+2x_2^2+3x_3^2+\ldots\leq1[/tex] so it is at most equal to 1 and this happens exactly at the point [tex](x_1,x_2,x_3,\ldots)=(1,0,0,\ldots).[/tex]
The absolute minimum at the boundary of this function happens when all the variables are equal to 0 except [tex]x_n=\frac{1}{\sqrt{n}}[/tex] and this minimum is equal to c=1/n. To see this notice that
[tex]nf=nx_1^2+nx_2^2+\cdots nx_n^2\geq x_1^2+2x_2^2+\cdots nx_n^2=1[/tex]
(the equality sign is because now we are on the boundary). We notice that nf is greater than or equal to 1 and the minimum of nf=1 (this implies the minimum for f to be 1/n) is attained exactly when [tex](x_1,x_2,\ldots,x_n)=(0,0,\ldots,\frac{1}{\sqrt{n}})[/tex].
So, finally, [tex]a+b+c=0+1+\frac{1}{n}=\frac{n+1}{n}.[/tex]
