Given positives $$\large a_1, a_2, \cdots, a_{n – 1}, a_n$$ $(n \in \mathbb Z^+, n \ge 3)$. Prove that for all naturals $p$ and $q$ such that $p \ge q$, $$\large \sum_{i = 1}^n\frac{\displaystyle \sum_{i' = 1}^na_{i'}^p – a_i^p}{\displaystyle \sum_{i' = 1}^na_{i'}^q – a_i^q} \le n \cdot \frac{\displaystyle \sum_{i = 1}^na_i^p}{\displaystyle \sum_{i = 1}^na_i^q}$$
We have that $$\sum_{i = 1}^n\frac{\displaystyle \sum_{i' = 1}^na_{i'}^p – a_i^p}{\displaystyle \sum_{i' = 1}^na_{i'}^q – a_i^q} \le n \cdot \frac{\displaystyle \sum_{i = 1}^na_i^p}{\displaystyle \sum_{i = 1}^na_i^q} \iff \sum_{i = 1}^na_i^q \cdot \sum_{i = 1}^n\frac{\displaystyle \sum_{i' = 1}^na_{i'}^p – a_i^p}{\displaystyle \sum_{i' = 1}^na_{i'}^q – a_i^q} \le n \cdot \sum_{i = 1}^na_i^p$$
$$\iff \sum_{i = 1}^n\frac{a_i^q \cdot \left(\displaystyle \sum_{i' = 1}^na_{i'}^p – a_i^p\right)}{\displaystyle \sum_{i' = 1}^na_{i'}^q – a_i^q} \le \sum_{i = 1}^na_i^p \iff \sum_{i = 1}^n\left[a_i^p – \frac{a_i^q \cdot \left(\displaystyle \sum_{i' = 1}^na_{i'}^p – a_i^p\right)}{\displaystyle \sum_{i' = 1}^na_{i'}^q – a_i^q}\right] \ge 0$$
$$\iff \sum_{i = 1}^n\frac{\displaystyle a_i^q \cdot \left(a_i^{p – q}\sum_{i' = 1}^na_{i'}^q – \sum_{i' = 1}^na_{i'}^p\right)}{\displaystyle \sum_{i' = 1}^na_{i'}^q – a_i^q} \ge 0 \iff \sum_{i = 1}^n\frac{\displaystyle a_i^q\sum_{i' = 1}^n[a_{i'}^q \cdot (a_i^{p – q} – a_{i'}^{p – q})]}{\displaystyle \sum_{i' = 1}^na_{i'}^q – a_i^q} \ge 0$$
$$\iff \sum_{i = 1}^n\sum_{i' = 1}^n\left[(a_ia_{i'})^q(a_i^{p – q} – a_{i'}^{p – q}) \cdot \left(\frac{1}{\displaystyle \sum_{i' = 1}^na_{i'}^q – a_i^q} – \frac{1}{\displaystyle \sum_{i' = 1}^na_i^q – a_{i'}^q}\right)\right] \ge 0$$
$$\iff \sum_{i = 1}^n\sum_{i' = 1}^n\left[\frac{(a_i^pa_{i'}^q – a_{i'}^pa_i^q)(a_i^q – a_{i'}^q)}{\left(\displaystyle \sum_{i' = 1}^na_{i'}^q – a_i^q\right) \cdot \left(\displaystyle \sum_{i' = 1}^na_i^q – a_{i'}^q\right)}\right] \ge 0$$
$$\iff \sum_{i = 1}^n\sum_{i' = 1}^n\left[\frac{(a_i^{p + q}a_{i'}^q + a_{i'}^{p + q}a_i^q) – (a_i^{2q}a_{i'}^p + a_{i'}^{2q}a_i^p)}{\left(\displaystyle \sum_{i' = 1}^na_{i'}^q – a_i^q\right) \cdot \left(\displaystyle \sum_{i' = 1}^na_i^q – a_{i'}^q\right)}\right] \ge 0$$
Then I don't know what to do next. Great.
Best Answer
Let $\sum\limits_{k=1}^na_k^p=t^p$, $\sum\limits_{k=1}^na_k^q=s^q$ and $x^{p-q}=\dfrac{t^p}{s^q},$ where $t,$ $s$ and $x$ are positives.
Thus, since by Karamata, $s^q>x^q$ and $\sum\limits_{k=1}^n(a_k^p-x^{p-q}a_k^q)=0,$ we obtain: $$\frac{n\sum\limits_{k = 1}^na_i^p}{\sum\limits_{k = 1}^na_i^q}-\sum_{k=1}^n\frac{\sum\limits_{i\neq k}a_i^p}{\sum\limits_{i\neq k}a_i^q}=\sum_{k=1}^n\left(\frac{t^p}{s^q}-\frac{t^p-a_k^p}{s^q-a_k^q}\right)=\frac{1}{s^q}\sum_{k=1}^n\frac{s^qa_k^p-t^pa_k^q}{s^q-a_k^q}=\sum_{k=1}^n\frac{a_k^p-x^{p-q}a_k^q}{s^q-a_k^q}$$ $$=\sum_{k=1}^n\left(\frac{a_k^p-x^{p-q}a_k^q}{s^q-a_k^q}-\frac{a_k^p-x^{p-q}a_k^q}{s^q-x^q}\right)=\sum_{k=1}^n\frac{a_k^q(a_k^{p-q}-x^{p-q})(a_k^q-x^q)}{(s^q-a_k^q)(s^q-x^q)}\geq0.$$ We see that our inequality is true for any positives $p$ and $q$ such that $p\geq q$.