Re: IMO LL Inequality problem
<font color="white">Assume the numbers are in decreasing order a1>=a2>=a3...
Probably a prettier solution, and this is a little hand-wavy, but: For a given S, when all but 2 numbers are held constant, varying the other two numbers has the following property (shown easily for a1>a2 and 0<x<a2 and 0<x<(a1-a2) by making the expressions a1+x/(S-(a1+x)) and a2-x/(S-(a2-x)) (or a1-x and a2+x)), increasing the larger number increases the value of the expression, and increasing the smaller number decreases the value of the expression. Given a constant S, minimizing the function requires that no number be bigger than any other number (otherwise it could be reduced by increasing the smaller number and decreasing the larger number), therefore all numbers must be the same. When this is true, a1=S/n, and the expression is equal to n*(S/n)/(S-S/n)=n/n-1.
Maximizing the function is similar. The biggest number must always be increased, but subject to the limit that a1<=a2+a3+..+an because these are sides of an n-gon (extension of the triangle inequality a1<=a2+a3). So increasing a1 up to S/2 at the expense of the other numbers (in reverse a_n, a_n-1, etc. order) will increase the value of function. Next, holding a1=S/2 constant, increasing a2 to S/2 at the expense of the remaining numbers will increase the function. So we get Approaching S/2, approaching S/2, approaching 0, approaching 0.. as the numbers. The function is then maximized at approaching (S/2)/(S-S/2) + (S/2)/(S-S/2) + 0/S +0/S.... = 1+1+0... = 2.</font>
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