max of $e$ with $a+b+c+d+e=8$ and $a^2+b^2+c^2+d^2+e^2=16$ [closed]

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 $e\geq 0$.

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 $a=b=c=d=2$.