Show that $\sum_{i=1}^{n}a_{i}^{n} \geq \sum_{i=1}^{n}b_{i}^{n}$ Conjecture: Let $a_{1}, a_{2},...,a_{n},b_{1},b_{2},...,b_{n}&gt;0$ be such that

$$\begin{align*}
\sum_{i=1}^{n}a_{i}\,&amp;\geq\,\sum_{i=1}^{n}b_{i}\,;\\
\sum_{i&lt;j}a_{i}a_{j}\,&amp;\geq\,\sum_{i&lt;j}b_{i}b_{j}\,;\\
\sum_{i\neq j\neq k}a_{i}a_{j} a_{k}\,&amp;\geq\,\sum_{i\neq j\neq k}b_{i}b_{j} b_{k}\,;\\
&amp;\ \ \vdots\\
a_{1}a_{2}\cdots a_{n}\,&amp;=\,b_{1}b_{2}\cdots b_{n}\,.
\end{align*}$$

It follows that

$$\sum_{i=1}^{n}a_{i}^{n} \geq \sum_{i=1}^{n}b_{i}^{n}$$

Examples:


$n = 2$: $a_{1}+a_{2} \geq b_{1}+b_{2},a_{1}a_{2} = b_{1}b_{2}$ then $a_{1}^2+a_{2}^2\geq b_{1}^2+b_{2}^2$.
$n = 3$: $a_{1}+a_{2} +a_{3} \geq b_{1}+b_{2} +b_{3} ,a_{1}a_{2}a_{3} = b_{1}b_{2}b_{3}$ then $a_{1}^3+a_{2}^3+a_{3}^3\geq b_{1}^3+b_{2}^3+b_{3}^3$.


Note: This is a repost of a question posted earlier today, that have been deleted by the author without reason. Since I put some-tough not too much-effort in the answer, I decided to repost it.
This can be proven easily by taking advantage of the AM-GM inequality:

$$\sum_{i=1}^{n}{a_i^n } \geq \sum_{i=1}^{n}{b_i^n } \Leftrightarrow \sum_{i=1}^{n}{a_i^n } \geq n\cdot \sqrt[n]{\prod_{i=1}^{n}b_{i}^{n}} |\cdot \frac{1}{n} \overset{\prod_{i=1}^{n}a_{i}}{\rightarrow}\frac{1}{n}\cdot \sum_{i=1}^{n}{a_i^n } \geq \sqrt[n]{\prod_{i=1}^{n}a_{i}^{n}}$$
This can be proven easily by taking advantage of the AM-GM inequality:

$$\sum_{i=1}^{n}{a_i^n } \geq \sum_{i=1}^{n}{b_i^n } \Leftrightarrow \sum_{i=1}^{n}{a_i^n } \geq n\cdot \sqrt[n]{\prod_{i=1}^{n}b_{i}^{n}} |\cdot \frac{1}{n} \overset{\prod_{i=1}^{n}a_{i}}{\rightarrow}\frac{1}{n}\cdot \sum_{i=1}^{n}{a_i^n } \geq \sqrt[n]{\prod_{i=1}^{n}a_{i}^{n}}$$