<s>[INST] Why do we consider the lower bound in this Inequality? Given that $a,b,c,d,e$ are real numbers such that
$a+b+c+d+e=8$,
$a^2+b^2+c^2+d^2+e^2=16$.
Determine the maximum value of $e$.

Let $t$ be a real number. Now consider the expression $$ \sum_{cyc} (t-a)^2 = \sum_{cyc} a^2-2t \sum_{cyc} a 
 + 5t^2$$
Now obviously $$(t-a)^2+(t-b)^2+(t-c)^2+(t-d)^2 \ge 0 $$ with equality at $a = b= c = d = t$ so $(t-e)^2 \le \sum_{cyc} a^2-2t \sum_{cyc} a + 5t^2$ or equivalently $e \le \sqrt{16-16t+5t^2} + t = f(t)$.
Now $f'(t) = 1+ \frac{5t-8}{16-16t+5t^2}$.
The resultant quadratic is easily solved for $t =6/5$ and $t = 2$, with the latter being an extraneous root introduced by squaring. The largest
possible $e $ and greatest lower bound of $f(t)$ is then $f(6/5) = 16/5$ , which occurs when
$a = b = c = d = 6/5 $ and $e = 16/5$ .

I had a small confusion here, why does the maximum possible value of $e$ where $f(t)$ is minimum, if $e≤f(t)$? Shouldn't it have been the maximum possible value of $f(t)$ in the possible interval of $t$ that would be allowed by the constraints, which here would just be at the endpoints since $f(t)$ has no maxima where it's derivative is $=0$.
In particular, from the second condition I am getting $-4≤e≤4$. But from first constraint, when $a=b=c=d=r$, $e=8-4t$ and thus equality can only hold when $t≤3$.
So why not just look at the values of $e$ at the endpoints of $t$ in this interval (ie at $3$ itself)?
 [/INST] Because if $e>\min\limits_{t}f$ so for $t=\frac{6}{5}$ and $a=b=c=d=\frac{6}{5}$ we'll obtain a contradiction.
Another way.
By C-S
$$16=a^2+b^2+c^2+d^2+e^2\geq\frac{1}{4}(a+b+c+d)^2+e^2=\frac{1}{4}(8-e)^2+e^2,$$ which gives $0\leq e\leq\frac{16}{5}.$
The equality occurs for $a=b=c=d$, id est occurs, which says that $\frac{16}{5}$ is a maximal value.
We got also that $0$ is a minimal value of $e$.
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