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

Apply the following identity involving an orthogonal array of sign patterns:
$4(a^2+b^2+c^2+d^2)=(a+b+c+d)^2+(a-b-c+d)^2+(a-b+c-d)^2+(a+b-c-d)^2$
Put in the given values and simplify giving
$3=(a-b-c+d)^2+(a-b+c-d)^2+(a+b-c-d)^2$
The condition $a\ge b\ge c\ge d>0$ forces $a+b-c-d$ to have an absolute value greater than or equal the other two squared terms on the right side, forcing $a+b-c-d\ge 1$.  This together with $a+b+c+d=9$ then guarantees that $a+b\ge 5$.
This does not assure that $5$ is the true minimum.  We have to be able to reach this value.  Render the following which satisfies the last equation and all squares the right side being equal:
$a-b-c+d=\pm 1, a-b+c-d=a+b-c-d=1$
With either choice of sign in the first relation, combine with the given equality $a+b+c+d=9$ to form a linear system with one of the two solutions
$a=3,b=c=d=2$ if $a-b-c+d=1$
$a=b=c=5/2, d=3/2$ if $a-b-c+d=-1$
Both of these realize $a+b=5$ making that the true minimum.
