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 $d\leq c\leq b\leq a\leq d+1$

Your solution is right and very nice!
The equality occurs for $x=4$ and $$(18x-x^2-2y-39)(2y-x^2)=0.$$
For $2y-x^2=0$ we obtain $y=8$, $c=d=2$, $a=3$ and $b=2$.
For $18x-x^2-2y-39=0$ we obtain $y=8.5$, $c=2.5$, $d=1.5$ and $a=b=2.5.$
We got all cases of equality occurring, as you said.
