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With ordered 4 real values satisfying the constraint as below, find the possible range of the largest one (i.e., the allowed interval for the variable $a$):

Let $a \geq b \geq c \geq d$ be reals satisfying: $$ 9\sum_{\mbox{cyc}} a^2 +10(ab+ ac+ad+bc+bd+cd) = 16.$$ Find the minimum and the maximum possible values for $a$.

I wrote the second term as $5 [(∑x_i)^2−∑x_i^2]$. Then, arrange the equality with variables sum and sum of squares to look like equation of an ellipse. From there, each term is bounded in absolute value (by values of the ellipse semi-axes). Then I bound for example $4d<a+b+c+d<4a$ and get some bounds for $d$. Note that if set $(a,b,c,d)$ is a solution, also the set $(−d,−c,−b,−a)$ is a solution.

Chip
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If you don't come up with a brighter idea, you can use Lagrange multipliers. Both the minimum and the maximum value of $a$ will come from stationary points of $$ L(a,b,c,d,\lambda) = a - \lambda(9(a^2+b^2+c^2+d^2)+ 10(ab+ac+ad+bc+bd+cd)-16). $$

The system you need to solve isn't too bad. You'll get to $a \in \left[-\sqrt{19/6},\sqrt{19/6} \right]$.

Note: If you reproduce the process and get the min/max values of the other variables, you'll see that they all have the same bounds, the given condition defines a 3-dimensional manifold $M \subset \mathbb{R}^4$ satisfying $M\subset \left[-\sqrt{19/6},\sqrt{19/6} \right]^4$

PierreCarre
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  • Pierre, that's nice. Using the Lagrange multiplier like that. Still, is it showing that the extremum obtained is a maximum / minimum simple / easy to see ? – Chip Oct 25 '23 at 13:21
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    @Chip that is a good point. In this case it is very easy actually to show that we are getting global minima/maxima. 1) Because of the regularity of the admissible region and of fact that the Jacobian of the restriction has maximum rank (1), any local minima/maxima are stationary point of the Lagrangian function. 2) Since the admissible region is compact and $f(a,b,c,d)=a$ is continuous, it will have a global maximum and minimum. Hence, the global max/min will be among the stationary points (there are exactly 2 stationary points) – PierreCarre Oct 25 '23 at 13:54