3

Let $a$, $b$, $c$, $d$, $e$ be real numbers such that $$ab+bc+cd+de+ea=20,\\ac+bd+ce+da+eb=22.$$ Find, as the root of a polynomial, the minimum value of $a^2+b^2+c^2+d^2+e^2$.

The numeric value of this minimum seems to be about $23.2359$.

I also tried Lagrange Multipliers but it didn't work.

Perhaps we could use some substitutions to eliminate the condition.

Update. I have proven that the problem is equivalent to (but no having the same answer as) the problem of minimizing $$\frac{(a-b)^2+(b-c)^2+(c-d)^2+(d-e)^2+(e-a)^2}{ac+bd+ce+da+eb-ab-bc-cd-de-ea}.$$

Update River Li finding the answer "$21+\sqrt5$" enabled me to produce the following natural solution that explores the essence of this problem (at least I think it does)

Here is the full process of my thinking.

We are to prove that $\sum\limits_{\rm cyc}a^2\ge21+\sqrt5$, or (by the first condition) $$(a-b)^2+(b-c)^2+(c-d)^2+(d-e)^2+(e-a)^2\ge2+2\sqrt5.\tag1$$ This, perceptually, means that the variables $a\dots$, $e$ must not stick too close to each other. So we will observe their differences by subtracting the conditions: $$a(b-c)+b(c-d)+c(d-e)+d(e-a)+e(a-b)=2.\tag2$$ Notice that if we increase each variable by $t\in\Bbb R$, then $(1)$ and $(2)$ are not affected. This means that given arbitrary numbers $A$, $B$, $C$, $D$, $E$ such that $$A(B-C)+B(C-D)+C(D-E)+D(E-A)+E(A-B)=2,$$ we are likely to be able to add $t\in\Bbb R$ to each of them so that the result $A'=A+t$ and similar satisfy $$A'B'+B'C'+C'D"+D'E'+E'A'=20.$$ Then we must also have $A'C'+\cdots+E'B'=22$. Moreover, as we just said, the addition doesn't affect $(1)$. Therefore: the numbers $22$ and $20$ are inessential. What's vital is that $22-20=\bf\color{red}2$. So the problem becomes:

Given real numbers $a$, $b$, $c$, $d$, $e$ such that $$a(b-c)+b(c-d)+c(d-e)+d(e-a)+e(a-b)=2,$$ prove that $\sum\limits_{\rm cyc}(a-b)^2\ge2+2\sqrt5$.

The proof is straightforward. It suffices to show that $$\sum_{\rm cyc}(a-b)^2\ge\left(1+\sqrt5\right)\sum_{\rm cyc}a(b-c),\tag3$$ which is easily true by Lagrangian SOS method.

Now we expand $(3)$, and the result is equivalent to River Li's identity!

youthdoo
  • 3,663
  • 1
    Is this problem a constest math problem? If yes, we can hope for a nice solution. If no, I doubt whether an analytical solution exists. – NN2 Mar 26 '24 at 08:48
  • @NN2 It is not a contest problem. – youthdoo Mar 26 '24 at 08:56
  • 1
    "As a root of a polynomial" I doubt you have come up with this yourself. What is the source? – D S Mar 26 '24 at 12:10
  • 1
    @Downvoters: Why did you put $-1$? I don't want to believe because you cannot solve the problem so explain your reasons please. – Ataulfo Mar 26 '24 at 15:06
  • 1
    @DS It is well known that any sort of "best constant" for algebraic inequalities is an algebraic number, although it might not have surd form. Hence I asked for a polynomial. – youthdoo Mar 26 '24 at 16:18
  • OP, you are lucky that this question can have a nice solution (me too, I learn a new technique). I give you an upvote to compensate the two downvotes you got. :) – NN2 Mar 26 '24 at 16:23
  • It is pertinent, I think, to remark that if $f(x)=0$ where $f$ is the polynomial having the roots $a,b,c,d,e$ then the coefficient of $x^3$ is $42$. – Ataulfo Mar 26 '24 at 16:30
  • 1
    I'm guessing that the downvotes are related to the post being considered low-quality. Esp for these inequality questions, just stating the problem and "I was unsuccessful" is considered insufficient, even though personally I think it's hard to provide much more context. That update might help substantiate that you've been thinking about this, esp if additional details were provided. – Calvin Lin Mar 26 '24 at 23:24

1 Answers1

4

Remark. Use the trick in my answer.

The minimum is $21 + \sqrt{5} \approx 23.23606798$.

Proof.

We have \begin{align*} &a^2 + b^2 + c^2 + d^2 + e^2 - (21 + \sqrt{5})\\[6pt] ={}& \frac{1 - \sqrt{5}}{2}(ab + bc + cd + de + ea - 20) + \frac{1 + \sqrt{5}}{2}(ac + bd + ce + da + eb -22)\\[6pt] &\qquad + \frac{3 - \sqrt{5}}{32}(b\sqrt{5} + c\sqrt{5} -2e\sqrt{5} - 2a + 3b + 3c - 2d - 2e)^2\\[6pt] &\qquad + \frac{5 + \sqrt{5}}{32}(b\sqrt{5} - c\sqrt{5} + 2a - b + c - 2d)^2\\[6pt] \ge{}& 0 \end{align*} with equality if \begin{align} &a = d = \frac15\,\sqrt {110+4\,\sqrt {135+80\,\sqrt {5}}+4\,\sqrt {5}},\\ &b = c = \frac15\,\sqrt {100-2\,\sqrt {-510+310\,\sqrt {5}}+4\,\sqrt {5}},\\ &e = -\frac{ab + bc + cd - 20}{a+d}. \end{align}

River Li
  • 49,125
  • +1. But how do you guess the coefficients $(1\pm \sqrt 5)/2$, $\frac{3 - \sqrt{5}}{32}$,... – NN2 Mar 26 '24 at 14:18
  • 1
    @NN2 Not guess. You can find constants $M, p, q$ such that $a^2 + b^2 + c^2 + d^2 + e^2 - M - p(ab + bc + cd + de + ea - 20) - q(ac + bd + ce + da + eb -22)$ is something like completing the squares. The idea is similar to this – River Li Mar 26 '24 at 14:26
  • 1
    I like this technique! – NN2 Mar 26 '24 at 14:31
  • Interesting. Is the problem of finding $M$, $p$, $q$ such that the rest is a sum of squares what we call "SDP problem"? – youthdoo Mar 26 '24 at 16:28
  • @youthdoo Yes. For quadratic function here, I think it is completing the squares in variables $e, \cdots$, one at a one. – River Li Mar 26 '24 at 22:10
  • I have a method that can smoothly determine $M$, $p$, $q$, and it is based on the "update" in the question! – youthdoo Mar 27 '24 at 09:58
  • 1
    @youthdoo Your proof in the update is very nice. – River Li Mar 27 '24 at 14:11