Determine the real numbers $a, b, c, d \in [1,3]$, knowing that the relation $(a + b + c + d)^2 = 3(a^2 + b^2 +c^2 + d^2)$.

Question

the question

Determine the real numbers $a, b, c, d \in [1,3]$, knowing that the relation $(a + b + c + d)^2 = 3(a^2 + b^2 +c^2 + d^2)$.

my idea

$(a+b+c+d)^2=a^2+b^2+c^2+d^2+2(ab+ac+ad+bc+bd+cd)$

$=> 2(ab+ac+ad+bc+bd+cd)=2(a^2 + b^2 +c^2 + d^2)=> ab+ac+ad+bc+bd+cd=a^2 + b^2 +c^2 + d^2$

From here I've been trying to get it to a form where 0 will equal the product of some numbers but I didn't get to anything helpful.

I hope one of you can help me! Thank you!

Answer 1

For $x \in \{a,b,c,d \}$, we have: $$(x-1)(x-3) \le 0 \iff x^2 \le 4x-3 \tag{1}$$ Let us denote $S = a+b+c+d$, from $(1)$ and the assumption, we deduce that $$\begin{align} &S^2 = 3 \sum_{\text{sym}}a^2 \le3 \left( \sum_{\text{sym}}(4a-3) \right) = 12S-36\\ \iff &(S-6)^2\le 0\\ \iff &S = 6 \tag{2} \end{align}$$

So, $(a,b,c,d)$ must satisfy the equality of $(1)$ and the equality of $(2)$.

As we cannot have $2$ variables having value equal to $3$, it's easy to deduce that the largest variable must be equal to $3$ and the 3 smaller variables receive value of $1$.

Answer 2

COMMENT.-Since $(3r+x)^2=3(3r^2+x^2)\iff r=\dfrac{x}{3}$ the only solution with three equal variables is $(a,b,c,d)=(1,1,1,3)$ and permutations since $\dfrac13$ and $\dfrac23$ are not allowed. Astonishing it would be the only solution. I neither affirm nor deny this statement in this simple comment (I suspect this is so).

Put for $a,b,c$ very small among allowed values, for example $(a,b,c,d)=(1,1.01,1.01,x)$ which gives the equation $(3.02+x)^2=3(3.0402+x^2)$ whose solutions are $x=0$ and $x\gt3$ and for greater values the solutions wil be not allowed. I think that what I suspect is true.

Answer 3

HINT:

Consider the function $$3(a^2 + b^2 +c^2 + d^2)-(a+b+c+d)^2$$ It is strictly convex in each variable separately ( being quadratic with leading coefficient $>0$). Such a function, on a product of closed intervals, achieves its maximum only at a vertex.

In this case the values at vertices are

$$3(1+1+1+1)-(1+1+1+1)^2=-4 \\ 3(1+1+1+9)-(1+1+1+3)^2= 0\\ 3(1+1+9+9)-(1+1+3+3)^2=-4\\ 3(9+9+9+9)-(3+3+3+3)^2=-36$$

We conclude that $\{a,b,c,d\} = \{1,1,1,3\}$.

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$\bf{Added:}$ Below explaining myself the neat solution of @NN2:

We have

$$(a+b+c+d)^2-3(a^2+b^2+c^2+d^2) =\\ =(a+b+c+d-6)^2 + 3 \sum (3-a)(a-1) \ge 0$$ with equality if and only if $a$, $b$, $c$, $d\in \{1,3\}$ and $a+b+c+d=6$, so one of them is $3$ and the others equal $1$.