Minimizing $a^2+b^2+c^2+d^2+e^2$ for real values satisfying $ab + bc + cd + de + ea = 10$ and $ac + bd + ce + da + eb = 11$

Question

Suppose $a, b, c, d, e$ are real numbers such that:
$$ab + bc + cd + de + ea = 10$$ $$ac + bd + ce + da + eb = 11$$ Find the minimum value of $$a^2 + b^2 + c^2 + d^2 + e^2$$

I started this question with

$$\begin{align} a^2 + b^2 + c^2 + d^2 + e^2 &= (a + b + c + d + e)^2 \\ &- 2(ab + bc + cd + de + ea + ac + bd + ce + da + eb) \end{align}$$

And then, $$y = a + b + c + d + e$$ $$a^2 + b^2 + c^2 + d^2 + e^2 = y^2 - 42$$

To minimize it, I would have to minimize $y$, which is where I got stuck

I saw a sort-of similar question with 4 variates, but that method didn’t work (or idk how to apply it).

If you could continue upon how to solve it from where I got stuck, please do so. Also, any other methods are welcome too.

Edit 1: I think inequalities like Cauchy, Trivial etc. measure the lower bound, not minimum value (correct me if I am wrong) so using them does not give the desired result.

$y^2$ cannot be smaller than 0 so you need to show that $y=0$ is possible. — garondal, Mar 25 '25 at 19:35
@garondal That would mean $a^2 + b^2 + c^2 + d^2 + e^2 = - 42$. Seriously? — , Mar 25 '25 at 19:43
As requested by the contest-math tag, please identify which contest this comes from so we can confirm it's not an open contest. Thanks. — Robert Shore, Mar 25 '25 at 22:49
@MathStackexchangeIsMarvellous (Superficial answer to why) If you click on the "Close(2)", you can see what options were chosen. In this case, both said "missing context". — Calvin Lin, Mar 25 '25 at 23:46
@RobertShore This is from a handout, i think? I don’t think this in particular was asked in a contest but similar questions have been asked (https://math.stackexchange.com/questions/3014786/smallest-value-of-a2-b2-c2-d2-given-values-for-abcd-ac) — Pradeep, Mar 26 '25 at 03:57
@CalvinLin Perhaps related to this question which you commented on. — River Li, Mar 26 '25 at 07:58
Mathematica answers $\frac{1}{2} \left(\sqrt{5}+21\right),a\to -\frac{1}{5} \sqrt{2 \sqrt{5}+2 \sqrt{80 \sqrt{5}+135}+55},\dots$ There is a lot of optimal solutions. — user64494, Mar 26 '25 at 10:19
This question is similar to: Find minimum of $a^2+\cdots+e^2$ under $2$ quadratic constraints. If you believe it’s different, please [edit] the question, make it clear how it’s different and/or how the answers on that question are not helpful for your problem. — Calvin Lin, Mar 26 '25 at 12:19

score 2 · Accepted Answer · answered Mar 26 '25 at 11:37

Do the substitution: $$a=a_0/\sqrt{2},b=b_0/\sqrt{2},c=c_0/\sqrt{2},d=d_0/\sqrt{2},e=e_0/\sqrt{2}$$ $$\implies a_0b_0+b_0c_0+c_0d_0+d_0e_0+e_0a_0=20,$$ $$a_0c_0+b_0d_0+c_0e_0+d_0a_0+e_0b_0=22,$$ $$a^2+b^2+c^2+d^2+e^2=\frac{1}{2}\left(a_0^2+b_0^2+c_0^2+d_0^2+e_0^2\right)$$ Using the same technique to this answer by River Li, we get: $$a_0^2+b_0^2+c_0^2+d_0^2+e_0^2\ge 21+\sqrt{5}$$ $$\implies a^2+b^2+c^2+d^2+e^2\ge \frac{1}{2}\left(21+\sqrt{5}\right)$$

You should simply have closed this as a duplicate. – Calvin Lin Mar 26 '25 at 12:09 — Calvin Lin, Mar 26 '25 at 12:09

Minimizing $a^2+b^2+c^2+d^2+e^2$ for real values satisfying $ab + bc + cd + de + ea = 10$ and $ac + bd + ce + da + eb = 11$

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