Prove that:$\frac{1}{a^n}+\frac{1}{b^n}+\frac{1}{c^n}\geqslant a^n+b^n+c^n$ Find $n\in\mathbb{N}^+$
For all Positive real numbers $a,b,c$ sastifying $a+b+c=3$
$\frac{1}{a^n}+\frac{1}{b^n}+\frac{1}{c^n}\geqslant a^n+b^n+c^n$

For an upper bound, consider $(a,b,c)=\left(\frac{3}{2},\frac{3}{4},\frac{3}{4}\right)$. (I thought of this myself, but I'm going to make it Community Wiki as this is reflected in the AoPS page.)
We need to solve for the smallest $n$ such that:
$\left(\frac{2}{3}\right)^n+2\left(\frac{4}{3}\right)^n<\left(\frac{3}{2}\right)^n+2\left(\frac{3}{4}\right)^n$.
Intuitively, a smallest $n$ should exist as $2\left(\frac{4}{3}\right)^n<<\left(\frac{3}{2}\right)^n$ for sufficiently large $n$.
A quick check on Wolfram Alpha gives $n>5.63493$.
As such, this inequality can only hold for $n=1,2,3,4,5$, since $n$ is an integer.
As a side note, the way I see how this works is trying things of the form $(a,b,c)=(3-2x,x,x)$, where $0<x<1$, as we want to make something on the right-hand side of the inequality to become very big.
The essential condition required to get a bound for $n$ is $\frac{1}{x}<3-2x$, or $2x^2-3x+1<0$. Factorising, we have $(2x-1)(x-1)<0$, which means $1>x>\frac{1}{2}$.
