sqing wrote: et $n \in \mathbb{N}^+,$ $x_1,x_2,...,x_{n+1},p,q\in \mathbb{R}^+ $ , $p<q$ and $x^p_{n+1}>\sum_{i=1}^{n}x^p_{i}.$ Prove that $(1)x^q_{n+1}>\sum_{i=1}^{n}x^q_{i};$ $(2)\left(x^p_{n+1}-\sum_{i=1}^{n}x^p_{i}\right)^{\frac{1}{p}}<\left(x^q_{n+1}-\sum_{i=1}^{n}x^q_{i}\right)^{\frac{1}{q}}.$

@above Could you translate your picture in the first post into LaTeX (and delete the extra $\$$ ) like Quote: Let $n \in \mathbb{N}^+,$ $x_1,x_2,...,x_{n+1},p,q\in \mathbb{R}^+ $ , $p<q$ and $x^p_{n+1}>\sum_{i=1}^{n}x^p_{i}.$ Prove that $(1)x^q_{n+1}>\sum_{i=1}^{n}x^q_{i};$ $(2)\left(x^p_{n+1}-\sum_{i=1}^{n}x^p_{i}\right)^{\frac{1}{p}}<\left(x^q_{n+1}-\sum_{i=1}^{n}x^q_{i}\right)^{\frac{1}{q}}.$ Otherwise, if it is added into a collection, the picture will be invisible, and there will be an error when generating PDF. Thank you.

This is an elementary solution I found. First, a little lemma that would shorten the proof: Lemma: $a_1, a_2, \dots, a_n \in \mathbb{R}^{+}$ and $r>1$, then $(a_1+a_2+...+a_n)^r > a_1^r+a_2^r+...+a_n^r$. Proof: Let's proof this lemma by induction. Consider $f(x)=(a_1+x)^r-a_1^r-x^r$, since $f'(x)=r(a_1+x)^{r-1}-rx^{r-1}>0$, we have that $f$ is a strictly increasing function. Therefore $f(a_2) > f(0)=0$, which means $(a_1+a_2)^r>a_1^r+a_2^r.$ Assume that the lemma is true for $n-1$, then $$\left(\sum_{i=1}^{n}a_i\right)^r = ((a_1+a_2)+...+a_n)^r > (a_1+a_2)^r+a_3^r+...+a_n^r > \sum_{i=1}^{n}a_i^r,$$so by induction the lemma is correct. (1) $x_{n+1}^q=\left(x_{n+1}^p\right)^{\frac{q}{p}} > \left(\sum_{i=1}^{n}x_i^p\right)^{\frac{q}{p}} > \sum_{i=1}^{n}x_i^q$. (2) On the other hand, by our lemma, \begin{align*} \sum_{i=1}^{n}x_i^q + \left(x_{n+1}^p - \sum_{i=1}^{n}x_i^p \right)^{\frac{q}{p}} &= \sum_{i=1}^{n}(x_i^p)^{\frac{q}{p}} + \left(x_{n+1}^p - \sum_{i=1}^{n}x_i^p \right)^{\frac{q}{p}} \\ \sum_{i=1}^{n}x_i^q + \left(x_{n+1}^p - \sum_{i=1}^{n}x_i^p \right)^{\frac{q}{p}} &< \left(x_{n+1}^p\right)^{\frac{q}{p}} \\ \left(x_{n+1}^p - \sum_{i=1}^{n}x_i^p \right)^{\frac{q}{p}} &< x_{n+1}^q - \sum_{i=1}^{n}x_i^q \\ \left(x_{n+1}^p - \sum_{i=1}^{n}x_i^p \right)^{\frac{1}{p}} &< \left(x_{n+1}^q - \sum_{i=1}^{n}x_i^q \right)^{\frac{1}{p}}. \end{align*}

FishHeadTail wrote: This is an elementary solution I found. I basically had the same solution. But I find it a bit funny that you call it "elementary" since it uses analysis (while the previous solution by sqing is actually elementary!).

Well most steps are. And the lemma "looks" true too, doesn't it?

Yes, I mean the elementary proof "à la Sqing" of the inequality $(a+b)^r>a^r+b^r$ for $r>1$ (which is the key inductive step of your lemma) is just \[\left(\frac{a}{a+b}\right)^r+\left(\frac{b}{a+b}\right)^r<\frac{a}{a+b}+\frac{b}{a+b}=1.\]