find min of $a+b$ given sum of $a\geq b\geq c\geq d$ is 9 and square sum is 21 Suppose $a\geq b\geq c\geq d>0$ and all are real numbers, and $a+b+c+d=9,a^2+b^2+c^2+d^2=21$, how to find the minmum of $a+b$?
What I attempted:
I can show $b\geq 1.5$ and $a\leq 3$.for $r\geq 0$, I consider $\sum(a-r)^2=\sum a^2-2r\sum a+4r^2=21-18r+4r^2$, then $a\leq \sqrt{21-18r+4r^2}+r$ which implies $a\leq 3$. Also I guess the min shoul be 5 and there are two solutions: 3,2,2,2 and $2.5,2.5,2.5,1.5$. But I cannot prove it.
New attempts:


*

*$6-2a=(a-3)^2+\sum(b-2)^2$

*$2d-3=\sum(a-2.5)^2+(d-1.5)^2$
then $2(d-a)+3\geq 1$ which implies that $d\leq c\leq b\leq a\leq d+1$

Set $c+d=x,c^2+d^2=y$ then we have
$$\begin{cases}c=\frac{x+\sqrt{2y-x^2}}{2}\\ d=\frac{x-\sqrt{2y-x^2}}{2}\end{cases}$$
Also it is easy to find out that 
$$b=\frac{9-x-\sqrt{42-2y-81+18x-x^2}}{2},$$
Note that $b\geq c$  we have 
$$(9-2x)^2\geq 18x-2x^2-39+2\sqrt{(18x-x^2-2y-39)(2y-x^2)}$$
which also implie that
$$6x^2-54x+120\geq 2\sqrt{(18x-x^2-2y-39)(2y-x^2)}\geq 0$$
so $x\geq 5$ or $x\leq 4$ and since $c+d\leq 9/4$ so $x\leq 4$ which means
$$a+b=9-x\geq 5.$$
And "$=$" happens when $(a,b,c,d)=(3,2,2,2) \text{ or } (2.5,2.5,2.5,1.5)$.
