find min of $a+b$ given sum of $a\geq b\geq c\geq d$ is 9 and square sum is 21

Question

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:

1. $6-2a=(a-3)^2+\sum(b-2)^2$

2. $2d-3=\sum(a-2.5)^2+(d-1.5)^2$

then $2(d-a)+3\geq 1$ which implies that $d\leq c\leq b\leq a\leq d+1$

Answer 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.

Answer 2

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.

Answer 3

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 $(a,b,c,d)=(3,2,2,2) \text{ or } (2.5,2.5,2.5,1.5)$.