Find the supremum and infimum (if exist) of the set
$S = \{\frac{m}{n} + \frac{4n}{m} \mid m,n \in \Bbb N\}$ and $T = \{x:x^2+x+1>0\}$.
Attempt:
First, notice that $S = \{x \in \Bbb R \mid x \gt 4\}$.
For the set $S$, by $AM-GM$ inequality, we have
\begin{align*}
\frac{m}{n} + \frac{4n}{m} &\ge 2\sqrt{\frac{m}{n} \cdot \frac{4n}{m}} = 2\cdot \sqrt{4} = 2\cdot 2 = 4.
\end{align*}
Thus, $4$ is a lower bound of $S$. We claim that $\inf(S)=4$. Let $m$ be an any lower bound of $S$.
We'll show that $m \le 4$. For the sake of contradiction, suppose $m \gt 4$. Let $r$ be an arbitrary rational number such that $4 < r < m$. Then, we have $r \in S$ and $m \gt r$. This is impossible since $m$ is a lower bound of $S$. Hence, $m \le 4$ and therefore, $\inf(S) = 4$.
Next, set $n=1$ and $m \to \infty$. Then, $\frac{m}{n} + \frac{4n}{m} \to \infty$.
Hence, $\sup(s)$ is doesn't exist.
Am I true?
${}$
For the set $T$, notice that $x^2+x+1$ has two distinct complex roots. How to process next?
What if $x$ there is in real set? What if $x$ there is in complex set?
Thanks in advanced.
Best Answer
Once you showed that $4$ is a lower bound, all that needs to be demonstrated is that the lower bound is attained for some $m, n \in \mathbb N$ in order for $\inf S = 4$. This is easily accomplished since AM-GM implies that equality is attained when $$\frac{m}{n} = \frac{4n}{m},$$ or $m = 2n$. So $(m,n) = (2,1)$ gives the desired lower bound, hence this must be the greatest lower bound.
For $x^2 + x + 1$, simply complete the square. $$x^2 + x + 1 = (x+1/2)^2 + 3/4 \ge 3/4,$$ and so all real numbers satisfy the condition $x^2 + x + 1 > 0$. Thus I suspect you intended to write instead $$T = \{x^2 + x + 1 : x \in \mathbb R\},$$ in which case $\inf T = 3/4$ and $\sup T = \infty$.