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Let $s_k = x^k + y^k + z^k + t^k$. First we check that the denominator is nonnegative. By Holder, we know $s_4^{2/3}s_1^{1/3} \ge s_3$, rearranging that and using $s_1 = 1$ we see that the denominator …

The inequality is true for all $n$. First of all, we can simplify it a little - from Douglas Zare's comment, we can assume $a_0 = 1$, $a_n = -1$, and try to maximize the LHS by varying the $a_i$s. Si …

This suggests that we might be able to prove all of these inequalities by induction on $n, k$: if we know it holds for $n, k-1$, then to prove it for $n,k$ it suffices to check the inequality when $a_n …

Ok, I have a functional generalization of your Product-Sum conjecture using a very simple method. Let $f:\mathbb{R}\rightarrow \mathbb{R}$ be any function with a nonnegative $\binom{n}{2}$th derivati …

Your expression is secretly a divided difference for the function $x \mapsto x^p$. In general, divided differences for a function $f$ are defined inductively by $f[x_0] = f(x_0)$ and $f[x_0, ..., x_n] …

Karamata's paper: http://elib.mi.sanu.ac.rs/files/journals/publ/1/11.pdf A few related papers are listed out in this AoPS forum post: http://www.artofproblemsolving.com/Forum/viewtopic.php?p=123878&# …

Even the inequality $(x,z;x,z)^2 \le (x,x;z,z)(z,z;x,x)$ is false: Let $V = \mathbb{R}^2$, with basis $x,z$. Take $(x,x;x,x) = 100$, $(x,z;x,x)=0$, $(z,z;x,x)=1$, $(x,x;x,z)=0$, $(x,z;x,z)=50$, $(z,z …

Furthermore, these simple inequalities can always be proved directly by a divided difference computation. … Things get a bit more complex once we jump to $\mathcal{C}(*) \le k+2$ — I used to be very interested in these sorts of inequalities a long time ago, and I have an article Inequalities and higher order …

I'm going to define an exponential polynomial of degree $k$ as a function $f$ of the form $f(x) = \sum_{i=1}^k c_ie^{\alpha_ix}$ ($\alpha_i$s real). My first question is: is there an algorithm for c …