max of $e$ with $a+b+c+d+e=8$ and $a^2+b^2+c^2+d^2+e^2=16$ Given that a,b,c,d,e are real number such that:
$\begin{cases}
a+b+c+d+e=8\\
a^2+b^2+c^2+d^2+e^2=16
\end{cases}$
determine the maximun value of $e$.
I started like that :
$\max(e)=\max(8-a-b-c-d)=\min(a+b+c+d)$
but i don't kow how to carry on 
can you help please ?
 A: Physics (Classical Mechanics) Solution:
Consider a $1$-dimensional elastic collision of a particle $X$ of mass $4$ moving at velocity $2$ into a particle $E$ of mass $1$, initially at rest.  Due to this collision, $X$ breaks into $4$ smaller particles $A$, $B$, $C$, and $D$ (of course, we are ignoring their binding energy, which would lead to an inelastic collision) with identical mass $1$, with velocities $a$, $b$, $c$, and $d$, and $E$ attains a velocity $e$.  Let $T_e$ denote the total energy of the particles $A$, $B$, $C$, and $D$ as a function of $e$.  Since $T_e$ is at least the kinetic energy of the center-of-mass frame of the particles $A$, $B$, $C$, and $D$, $$T_e\geq \frac{1}{2}\cdot (1+1+1+1)\cdot \left(\frac{a+b+c+d}{4}\right)^2=\frac{(a+b+c+d)^2}{8}\,.$$
The equality case of the inequality above is when $a=b=c=d$ (i.e., when the particles $A$, $B$, $C$, and $D$ are at rest in their center-of-mass frame).  By the Conservation Law of Momentum, $$a+b+c+d=4\cdot 2-1\cdot e=8-e\,.$$  By the Conservation Law of Energy, $$T_e=\frac{1}{2}\cdot4\cdot 2^2-\frac{1}{2}\cdot 1\cdot e^2=\frac{16-e^2}{2}\,.$$
Hence, 
$$\frac{16-e^2}{2}=T_e\geq \frac{(a+b+c+d)^2}{8}=\frac{(8-e)^2}{8}\,,$$
whence $0\leq e\leq \frac{16}{5}$.  The maximum $e=\frac{16}{5}$ is attained iff $a=b=c=d=\frac{6}{5}$.  The minimum $e=0$ is attained iff $a=b=c=d=2$.
A: By the Cauchy-Schwarz inequality,
$$ (8-e)^2 = (a+b+c+d)^2 \leq 4(a^2+b^2+c^2+d^2) = 4(16-e^2) $$
from which it follows that $e\leq \color{red}{\frac{16}{5}}$. Now it is enough to show that the inequality holds as an equality for some $(a,b,c,d,e)\in\mathbb{R}^5$, pretty easy. In the same way you may also show that $e\geq 0$.
A: The geometry is the same as in 3 dimensions, when taking the plane section $x+y+z = c$ of a sphere $x^2+y^2+z^2=C$.  Extrema of $z$ are when $x=y$.  The only remaining thing to understand is why the answer for the parameters given in the problem does not involve square roots.
Let $n$, which equals $4$ in the posted problem, be the number of variables other than the last one, which I will continue to call "$e$".
Max/min $e$ in $(n+1)$ dimensions occur at the two solutions of 
$na+e=c$ and $na^2+e^2=C$, so $(c-e)^2 = n(C - e^2)$ or $$(n+1)e^2 - 2ce + (c^2 - nC) = 0 .$$
To get a problem with extremal $e$ rational in the parameters, 
$c^2 - (n+1)(c^2 - nC^2) = n((n+1)C - c^2)$ must be a perfect square. 
There is no special perfect square expression that appears in the algebra, and the problem was just based on a choice of small integer values that lead to a square.
