# A tough inequality with a constraint?

Let $$x_1,$$ $$x_2,$$ $$x_3,$$ $$x_4,$$ $$x_5$$ be distinct positive integers such that $$x_1 + x_2 + x_3 + x_4 + x_5 = 100.$$ Compute the maximum value of the expression \begin{align*} &\frac{(x_2 x_5 + 1)(x_3 x_5 + 1)(x_4 x_5 + 1)}{(x_2 - x_1)(x_3 - x_1)(x_4 - x_1)} + \frac{(x_1 x_5 + 1)(x_3 x_5 + 1)(x_4 x_5 + 1)}{(x_1 - x_2)(x_3 - x_2)(x_4 - x_2)} \\ &\quad + \frac{(x_1 x_5 + 1)(x_2 x_5 + 1)(x_4 x_5 + 1)}{(x_1 - x_3)(x_2 - x_3)(x_4 - x_3)} + \frac{(x_1 x_5 + 1)(x_2 x_5 + 1)(x_3 x_5 + 1)}{(x_1 - x_4)(x_2 - x_4)(x_3 - x_4)}. \end{align*}

There is probably some clever insight that I haven't seen, any ideas?

• Maybe you could do a little computer simulation getting the maximum value. Using that, you could improve the question as: "Show that the maximum value of the expression (...) is obtained for the following values for $x_i$: (...)". – Andreas Apr 16 '19 at 21:28
• I got it. It resembles the lagrange interpolation formula, which can be cleverly applied here. – doingmath Apr 17 '19 at 1:23
• Is your sum equal to $x_5^3$? If one factors it out, one obtains the sum of the Lagrange polynomials for $x_1,...,x_4$ at the point $-1/x_5$. This sum is 1, of course. – Helmut Apr 19 '19 at 16:14

If the terms are factored out, the expression equals $$x_5^3$$ (see Helmut's comment, I checked it with Matlab).
Since we have that $$x_1,$$ $$x_2,$$ $$x_3,$$ $$x_4,$$ $$x_5$$ are distinct positive integers such that $$x_1 + x_2 + x_3 + x_4 + x_5 = 100$$, the maximum value of the expression will be obtained for the smallest possible distinct integers $$x_1,$$ $$x_2,$$ $$x_3,$$ $$x_4$$. Choose $$(1,2,3,4)$$ for these four values, then the maximum value of the expression is $$90^3 = 729000$$.
• +1. BTW, treat RHS as a cubic polynomial in $x_5$. Let's call it $f(x_5)$. Notice $f(-\frac{1}{x_i}) = -\frac{1}{x_i^3} = \left(-\frac{1}{x_i}\right)^3$ at $4 > 3$ points, $f(x)$ equals to $x^3$ as a polynomial. i.e. the terms do factored out. – achille hui Aug 25 '19 at 9:16