Minimum and maximum There are five real numbers $a,b,c,d,e$ such that
$$a + b + c + d + e = 7$$$$a^2+b^2+c^2+d^2+e^2 = 10$$
How can we find the maximum and minimum possible values of any one of the numbers ?
Source
 A: Maximum value that can be taken by one of the variables is $\frac{9}{5}$ and the minimum value is $1$.
This can be argued based on symmetry as follows (and I hope I have not overlooked anything in my symmetry argument).
Say we want to find the extreme values of $a$. By symmetry when $a$ reaches the extremum the remaining variables have to be equal. So when $a$ takes the extremum $b=c=d=e$.
Hence, $a + 4b = 7$ and $a^2 + 4b^2 = 10$ i.e. $a^2 + 4 (\frac{7-a}{4})^2 = 10$ 
i.e. $a^2 + \frac{(a-7)^2}{4} = 10$ i.e. $5a^2 -14a + 49 = 40 $ i.e. $5a^2 - 14a + 9 =0$.
Hence, $(5a-9)(a-1)=0$.
So, minimum is $a=1$ and maximum is $a = \frac{9}{5}$.
The round about way would be to use Lagrange multiplier technique and set up 7 equations in 7 unknowns and solve it.
A: So you want $x\in \mathbb{R}^5$ such that $\| x\|^2 = 10$  and $(x,1)=7$.  (Here 1 refers to the all 1's vector)
Geometrically, we are looking at a circle around the all 1's vector in $\mathbb{R}^5$, and we just need to find the extremes of the circle.
From the above discussion it should be clear that the maximum and minimum will be close to $\sqrt {2}$, but the question is how much can it deviate?
Now lets actually solve the problem:  Consider $a=b=c=d$ so our equations are in two variables, $$4a+e=7$$
$$4a^2+e^2=10$$
Solving $a=\frac{7-e}{4}$ and plugging into the second equation we get the quadratic equation:  $$5e^2-14e+9=0$$  Which factors nicely as $$(5e-9)(e-1)=0$$  So either $e=1$ or $e=1.8$.  I now claim that these are indeed the minimum and maximum values, but I will leave that for you to prove.
Hint:  The vector $x$ in $\mathbb {R}^n$ with $\|x\|\leq r$ that maximizes $(x,1)$ will be in the direction of $1$ (the all 1's vector).  In other words, maximizing $a+b+c+d$ given $a^2+b^2+c^2+d^2<10-1.8^2$ occurs when $a=b=c=d$.  Similarly for minimizing.
A: I just wanted to point out that if you don't want to appeal to symmetry there is a Lagrange multiplier argument which isn't too bad: Plugging in the first equation into the second, the last 4 variables satisfy the equation
$$(7 - b - c - d - e)^2 + b^2 + c^2 + d^2 + e^2 = 10$$
Suppose you want to maximize $b$ for example. Then the gradient of $(7 - b - c - d - e)^2 + b^2 + c^2 + d^2 + e^2$ has to be a multiple of the gradient of $b$, which means there is some $\lambda$ such that
$$-2(7 - b - c - d - e) + 2b = \lambda$$
$$-2(7 - b - c - d - e) + 2c = 0$$
$$-2(7 - b - c - d - e) + 2d = 0$$
$$-2(7 - b - c - d - e) + 2e = 0$$
But the last three equations are the same as $-2a + 2c = 0$, $-2a + 2d = 0$, and
$-2a + 2e = 0$, which lead to $a = c = d = e$. Thus the $4$ variables other than $b$ are the same and one can now proceed as in Sivaram's argument.
