I am a novice SE user, a toddler. I was notified that one of my edits, https://math.stackexchange.com/review/suggested-edits/1384537 , was rejected. I followed the guideline of editing; there were some grammatical errors needed to be fixed. I think that "being substantial" means "changing at leas...

$$\text{If } x \geq \frac{a}{b} \text{ and } x \geq \frac{c}{d}$$ for some $a, c \in \mathbb{R}_{\geq 0}$ and $b, d \in \mathbb{Z}_{\geq 1}$, is it true that $x \geq \frac{a + c}{b + d}$?

So here is the Question . Prove that for all $a,b,c > 0$ :- $$\frac{ab}{a^5 + b^5 + ab} + \frac{bc}{b^5 + c^5+ bc} + \frac{ca}{c^5 + a^5 + ca} \leq 1 : a,b,c \in \mathbb{R}; abc = 1$$ Can anyone help me how to prove this?

Prove that if $N \equiv 3 \pmod{10}$, where $N = n^{n+1}-(n+1)^n$, then $n \equiv 11 \pmod {20}$ ($n$: Natural number).

How can we show that $$ \sum\limits_{i=0}^{r-j} (-1)^{i+1} \binom{i+j}{i}\binom{r-l}{i+j-l}= \left\{ {\begin{array}{*{20}{c}} 0 \;\;\;\;\;\;\;\;\;\;\;\;\;\;\;{\ \ (j < r-l),} \\ {(-1)^{r-j-1}\binom{l}{r-j}\;\;\;(r-l\leq j\leq r).} \\ \end{array} } \right. $$ What kind of method can we use...

Given that the n-th root is defined as follows$$ {\sqrt[n]a} = a^{1/n}$$ Can we infer that n can be irrational, negative or even imaginary? Usually we mean the square root (n=2) and then the third root (n=3) but is there no reason to think that there could not be n=2.1 for example, and that root ...

How would I approach something like this?