a) $S_{k+1}S_1 \geq S_kS_2$ because $S_{k+1}S_1-S_kS_2= \sum x_ix_j(x_i^k+x_j^k-x_i^{k-1}x_j-x_j^{k-1}x_k)= \sum x_ix_j(x_i-x_j)^2(x_i^{k-2}+x_i^{k-3}x_j+...+x_j^{k-2})$ And so $S_{k+1} \geq \frac{S_kS_2}{S_1}>S_k$ b)$x_1=2, x_2=x_3=x_4=...x_{10}=x_{11}=\frac{1}{2}$ Then $S_1=7,S_2=\frac{13}{2}<S_1,S_3=\frac{37}{4}>S_1$ so $S_1, S_2, S_3, \ldots$ is not strictly decreasing.

(a) Let $f(t)=x_1^t+x_2^t+\ldots+x_n^t$ for $t\in [0,\infty)$ such that $S_k = f(k)$. Note that $f(t)$ is convex, for instance because $f''(t) = \log(x_1)^2 x_1 ^t + ... + \log(x_n)^2 x_n ^t \geq 0$. Therefore $f(k+1) - f(k) \geq f(k) - f(k-1)$. By induction, we get $f(k+1) - f(k) \geq f(2)-f(1) > 0$ for $k\geq 1$, which implies that the sequence $f(k)$ is strictly increasing. (b) Let $n=2$ and $x_1=1.01, x_2 = 0.5$. It is easy to verify that $S_1 = x_1 + x_2 > x_1^2 + x_2^2 = S_2$, however since $S_k \to \infty$ as $k\to \infty$, the sequence cannot be strictly decreasing.

(a) From Cauchy-Schwarz we have $S_k^2 \le S_{k-1}S_{k+1}$ and it follows immediately that $\frac{S_{k+1}}{S_k}$ is monotonically increasing, hence $S_{k+1}>S_k$ if $S_2>S_1$. (b) Almost anything works. Let's just take $n=2$ and $x_1=c, x_2=2c$. If $c>\frac{1}{2}$, we have $x_2>1$ and hence $S_k \to \infty$. On the other hand, $S_1=3c$ and $S_2=5c^2$ so for $c<\frac{3}{5}$ we will have $S_1>S_2$. Choosing any $c$ between $\frac{1}{2}$ and $\frac{3}{5}$ therefore wins, e.g. we can take $c=\frac{4}{7}$ and hence $x_1=\frac{4}{7}, x_2=\frac{8}{7}$.

The $a$ part only: As $x_i = 1$ contribute the same for both sides of an inequality of the form $S_{i+1} > S_i$ we can asume all $x_i$ are $x_i \neq 1$. As we can see the given inequality $S_1 < S_2$ turns into: $$\sum_{x_i > 1} x_i(x_i-1) > \sum_{x_i < 1} x_i(1-x_i)$$So now the inequality $S_{k+1} > S_k$ can be achieve by noticing: $$\sum_{x_i > 1} x_i^k(x_i-1) \geq \sum_{x_i > 1} x_i(x_i-1) > \sum_{x_i < 1} x_i(1-x_i) \geq \sum_{x_i < 1} x_i^k(1-x_i)$$

a) Lemma: $\dfrac{S_{k+2}}{S_{k+1}}>\dfrac{S_{k+1}}{S_k}$, $\forall k\in\mathbb{Z^*_+}$ Proof: We will prove by induction in $k$. (i) Base case: if you want, just do a particularization of the induction step, whatever. (ii) Hypothesis: Assume the lemma is valid for all $m\leq k$, i.e, $\dfrac{S_{k+2}}{S_{k+1}}>\dfrac{S_{k+1}}{S_k}>\dfrac{S_{k}}{S_{k-1}}>\dots >\dfrac{S_2}{S_1}$. (iii) Induction Step: We will prove that is also valid for $k+1$. So, we want: $\dfrac{S_{k+3}}{S_{k+2}}>\dfrac{S_{k+2}}{S_{k+1}}\iff S_{k+3}\cdot S_{k+1}>S_{k+2}^2\iff$ $$\sum_{i=1}^n X_i^{k+3}\cdot \sum_{i=1}^n X_i^{k+1}>\bigg(\sum_{i=1}^n X_i^{k+2}\bigg)^2\iff \sum_{i=1}^n X_i^{2k+4}+\sum_{i\neq j} X_i^{k+3}X_j^{k+1}>\sum_{i=1}^n X_i^{2k+4}+2\cdot\sum_{i\neq j} X_i^{k+2}X_j^{k+2}$$ $$\iff \sum X_i^{k+1}X_j^{k+1}(X_i^2+X_j^2)> 2\cdot \sum X_i^{k+2}X_j^{k+2}$$. By AM-GM, we're done. Therefore, in particular, we have that $\dfrac{S_{k+2}}{S_{k+1}}>\dfrac{S_2}{S_1}>1\implies S_{k+2}>S_{k+1}, \forall k$. b) Let $X_1= \dfrac{5}{4}$, $X_2= X_3= \dfrac{1}{4}$. So, $S_1= \dfrac{7}{4}$, $S_2= \dfrac{27}{16}$, $S_3= \dfrac{127}{64}$: $S_1=\dfrac{7}{4}>\dfrac{27}{16}=S_2$, $S_2= \dfrac{27}{16}<\dfrac{127}{64}= S_3.$