Bad sketch:

Bad solution:

How many marks would you lose if you forget to provide a construction?

Darn while looking at the problems after school I found the complex factorization but thought that trivial inequality couldn't work since I had never used the $\{x_i\}_{i=1}^4$ real condition.

Edit: looks like djmathman beat me to this one.

This is a truly horrible problem Using Vieta we get that the relation equals $a^2+b^2+c^2+d^2-2d-2ac+2d-2b+1 = (a-c)^2+(X)(X+2)+1$ where $X = d-b \le -5$. And so the maximum value is $16$ , achieved when the polynomial is $(x-1)^4$ ($b-d=6-1$).

I don't think you need the roots to be real. When is it used?

^When you use the trivial inequality, as it only holds for reals.

But a and c are given to be reals, so the trivial inequality holds anyway.

yeah real roots are not needed. Though otherwise it isn't clear that the desired product is real, so it makes sense that they would include the condition just for that.

Oh yeah, the product has to be real. But you don't need it for the proof. I was wondering why they had that condition...

brian22 wrote: Oh yeah, the product has to be real. But you don't need it for the proof. I was wondering why they had that condition... They wanted you to come up w/ an explicit example and demonstrated that such a set of roots actually existed.

So what I did is (assuming r, s, t, u to be the roots) show that (r^2 + 1)(s^2 + 1)(t^2 + 1)(u^2 + 1) >= (rs + rt + ru + st + su + tu - rstu - 1)^2 which is equivalent to (rst + rsu + rtu + stu - r - s - t - u)^2 >=0 which is obviously true. However in my bashy expansion steps I misused sym notation - for example I said "by Vieta's formula we have b = sum{sym}(rs) when in fact I want half of that. I was pressed for time so at the end of my proof I wrote "I misused sym notation but the proof should still be clear." All my methods were right, I merely notated wrong. Do you think I will be graded 7- (that is, be given a 5-7)?

Knightone wrote: brian22 wrote: Oh yeah, the product has to be real. But you don't need it for the proof. I was wondering why they had that condition... They wanted you to come up w/ an explicit example and demonstrated that such a set of roots actually existed. wait but I didn't show this. How much would I get marked down?

I would say 1 or 2 points. I'm not sure how much it matters, although it is pretty clear that you should show an equality case.

I'm not an experienced grader by any means, but I believe that if you don't show an equality case then you should get 2 out of 7 points, since well, you didn't actually prove that your answer was attainable, and thereby didn't completely solve the problem.

How do you think the notational mistake I wrote about a couple of posts above would be graded?

Alright, this problem was some real pain... Firstly expand the entire expression to get that,\[(x_1^2+1)(x_2^2+1)(x_3^2+1)(x_4^2+1)=(a-c)^2+((b-d)-1)^2,\]where we derive this form by expanding and then grouping terms, like $(x_1^2+x_2^2+x_3^2+x_4^2)=(x_1+x_2+x_3+x_4)^2-2(x_1x_2+x_1x_3+x_1x_4+x_2x_3+x_2x_4+x_3x_4)$, and similarly for the other terms too, then do some crazy amounts of Vieta spam to get the final form. Now clearly the minimum of $(a-c)^2+((b-d)-1)^2=16$ due to the given condition. Now we just show the construction. Take $P(x)=(x-1)^4=x^4-4x^3+6x^2-4x+1$ which has $a=c=-4$ and $b=d+5=1+5=6$ and we are done.

Assume the roots of the polynomial $P(x)=x^4+ax^3+bx^2+cx+d$ are $\alpha ,\beta ,\gamma ,\delta.$ we have $\boxed{b-d \ge 5}.$ we have to find the value of $$\boxed{(\alpha^2+1)(\beta^2+1)(\gamma^2+1)(\delta^2+1)=\prod_{cyc}{}[(\alpha+i)(\alpha-i)]_{min}}$$Putting the $i$ and $-i$ in given equation we get since they ask the minimum value then have $d \ge b-5\implies \boxed{d_{min}=b-5}$ $$\boxed{\prod_{cyc}{}(\alpha+i)=1-ai-b-ci+d}$$$$\boxed{\prod_{cyc}{}(\alpha-i)=1+ai-b-ci+d}$$Multiply these we get $$\prod_{cyc}{}(\alpha+i)(\alpha-i)=(1-b+d)^2+(a-c)^2\ge (5-1)^2=16.$$therefore we get $$\boxed{\prod_{cyc}{}(\alpha+i)(\alpha-i)\ge 16}$$is the correct answer!!!!! Vietas Relation!!!!!!!!!

Write \begin{align*} \prod_{n=1}^4(1+x_n^2)&=\prod_{n=1}^4(i-x_n)(-i-x_n)\\ &=P(i)P(-i)\\ &=(1-b+d)^2+(a-c)^2\\ &\ge\boxed{16}, \end{align*}with equality at $P(x)=(x-1)^4$ where $b-d=5$ and $x_1=x_2=x_3=x_4=1$. $\square$

$$\prod_{\text{cyc}}(x_{1}^{2}+1)=\prod_{\text{cyc}}(x_{1}+i)(x_{1}-i)=P(i)P(-i)$$$$P(i)P(-i)=(1-b+d+(c-a)i)(1-b+d-(c-a)i)=(1-(b-d))^2+(c-a)^2$$To minimize $(1-(b-d))^2$ let $b-d=5$ and set $c=a$ to get a minimum of $4^{2}=16$ with equality at $p(x)=(x-1)^4$

I claim that the minimum value is $\boxed{16}$, when $x_1 = x_2 = x_3 = x_4 = 1$. By Vieta's formulas we have that \begin{align*} \sum_{sym} x_1^2 &= a^2 - 2b \\ \sum_{sym} x_1^2x_2^2 &= b^2 - 2 \sum_{sym} x_1 \sum_{sym} x_1x_2x_3 = b^2 - 2ac \\ \sum_{sym} x_1^2x_2^2x_3^2 &= c^2 - 2bd \\ x_1^2x_2^2x_3^2x_4^2 &= d^2 \\ \end{align*}Therefore if we expand our desired product, we obtain \[(a - c)^2 + (b - d)(b - d - 2) + 1 \geq 0^2 + 5(3) + 1 = 16. \blacksquare\]

$(x_1^2+1)(x_2^2+1)(x_3^2+1)(x_4^2+1)=(x_1-i)(x_2-i)(x_3-i)(x_4-i)(x_1+i)(x_2+i)(x_3+i)(x_4+i)=P(i)P(-i)=((1-b+d)+i(-a+c))((1-b+d)-i(-a+c))=(1-b+d)^2+(-a+c)^2 \ge 16+(c-a)^2 \ge \boxed{16}$

We can rewrite the expression, we are trying to find the minimum of as: \begin{align*} \prod_{j=1}^{4}(x_j+i)(x_j-i)&=P(i)\cdot P(-i)\\ &=(1-ai-b+ci+d)(1+ai-b-ci+d)\\ &=(1-b-d)^2+(a-c)^2\\ &\geq (b+d-1)^2\\ &\geq 16\\ \end{align*}We therefore claim the minimum is $16$. For a construction, let $x_1=x_2=x_3=x_4=1$, resulting in $b-d=6-1\geq 5$, so it works.

Consider the polynomial $Q$ whose roots are the squares of that of $P$. It can be easily seen that $$Q(x) = x^4 + (2b-a^2)t^3 + (b^2 + 2d - 2ac)t^2 + (2bd - c^2)t + d^2.$$Putting in $x = -1$, $$Q(-1) = 1 + a^2 - 2b + b^2 + 2d - 2ac + c^2 -2bd + d^2$$$$=(a-c)^2 + (b-d-1)^2.$$Clearly $(a-c)^2 \ge 0$, and $(b-d-1)^2 \ge 4^2 = 16$. Therefore $(x_1^2+1)(x_2^2+1)(x_3^2+1)(x_4^2+1) \ge 16$, and since $P(x) = (x+1)^4$ works, the answer is in fact $\boxed{16}$. $\square$

We want to find the minimal value for $A = (x_1^2 + 1)(x_2^2 + 1)(x_3^2 + 1)(x_4^2 + 1) = x_1^2x_2^2x_3^2x_4^2 + x_1^2x_2^2x_3^2 + x_1^2x_2^2x_4^2 + x_1^2x_3^2x_4^2 + x_2^2x_3^2x_4^2 + x_1^2x_2^2 + x_1^2x_3^2 + x_1^2x_4^2 + x_2^2x_3^2 + x_2^2x_4^2 + x_3^2x_4^2 + x_1^2 + x_2^2 + x_3^2 + x_4^2 + 1$. Now by Vieta's we get that $x_1 + x_2 + x_3 + x_4 = -a$, $x_1x_2 + x_1x_3 + x_1x_4 + x_2x_3 + x_2x_4 + x_3x_4 = b$, $x_1x_2x_3 + x_1x_2x_4 + x_1x_3x_4 + x_2x_3x_4 = -c$, $x_1x_2x_3x_4 = d$. By all that we get $x_1^2 + x_2^2 + x_3^2 + x_4^2 = a^2 - 2b$, $x_1^2x_2^2 + x_1^2x_3^2 + x_1^2x_4^2 + x_2^2x_3^2 + x_2^2x_4^2 + x_3^2x_4^2 = b^2 - 2ac$, $x_1^2x_2^2x_3^2 + x_1^2x_2^2x_4^2 + x_1^2x_3^2x_4^2 + x_2^2x_3^2x_4^2 = c^2 - 2bd$, $x_1^2x_2^2x_3^2x_4^2 = d^2$. Now we get that $A = d^2 + c^2 - 2bd + b^2 - 2ac + a^2 - 2b + 1 = (a - c)^2 + (b - d)(b - d - 2) + 1 \geq 0^2 + 5.3 + 1 = 16$, which is achievable when $x_1 = x_2 = x_3 = x_4 = 1$ $\Rightarrow$ $A \geq 16$, and for A = 16 exists an example $\Rightarrow$ the answer is 16.

The answer is $16$, given by $P(x) = (x+1)^4$. Notice that$$(x_1^2+1)(x_2^2+1)(x_3^2+1)(x_4^2+1)=(x_1+i)(x_1-i)(x_2+i)(x_2-i)(x_2+i)(x_2-i)(x_2+i)(x_2-i).$$Clearly $P(x) = (x_1 - x)(x_2-x)(x_3-x)(x_4-x)$ so the RHS of the equation above is $P(i)P(-i)$. We have $P(i)P(-i) = (b-d-1)^2 + (a-c)^2 \ge 4^2 = 16.$ $\blacksquare$

Observe that $x^2+1=x^2-(-1)=x^2-i^2=(x-i)(x+i).$ Now, notice that:\[\prod_{k=1} ^4 (x_k-i)(x_k+i)=\prod_{k=1}^4(x_k-i)\cdot\prod_{k=1}^4(x_k+i).\]The definition of $P(x)$ is $(x-x_1)(x-x_2)(x-x_3)(x-x_4)$ where the leading coeffecient is $1$ since $P$ is a monic polynomial as given in the problem. Then, substituting in $i,$ we have:\[P(i)=(i-x_1)\cdots(i-x_4)=(-1)^4(x_1-i)\cdots(x_4-i).\]This is exactly our first product. Our second product can be found as follows:\[P(-i)=(-i-x_1)\cdots(-i-x_4)=(-1)^4(x_1+i)\cdots(x_4+i).\]Hence, what we want to find is $P(i)P(-i).$ But substituting in $i$ into our expressoin $x^4+ax^3+bx^2+cx+d$ gets us $P(i)=(1-b+d)+(c-a)i,$ and $P(-i)=(1-b+d)+(a-c)i.$ Hence:\[P(i)P(-i)=((1-b+d)+(c-a)i)((1-b+d)+(a-c)i)=(1-b+d)^2-(a-c)^2=(-1)^2(b-d-1)^2-(a-c)^2 \ge 4^2 -0^2=16\]where equality holds at $a=c$ and $b-d=5 \Rightarrow b=d+5.$ Hence, the minimum is $\boxed{16}.$ To finish this off we find a construction for this minimum. We know that $a=c$ and $b=d+5.$ Hence\[P(x)=x^4+cx^3+(d+5)x^2+cx+d.\]We set $d=1$ to get as much symmetry as possible within our polynomial. This leads to $x^4+cx^3+6x^2+cx+1.$ However, note that $\binom{4}{2}=6$ and $\binom{4}{0}=\binom{4}{4}=1,$ so with some wishful thinking this leads us to think about the binomial theorem. We can try $(x\pm 1)^4$ and realize that those solutions do work. Hence, $16$ is obtained when all of $x_i$ are equal to $1$ or all equal to $-1.$

We claim the minimum value that the product \[\prod_{k = 1}^{4} (x_k^2 + 1)\] can take is $16$. Rewrite the product as \[\prod_{k = 1}^{4} (x_k + i)(x_k - i) = \prod_{k = 1}^{4} (x_k + i) \prod_{k = 1}^{4} (x_k - i)\] where $i = \sqrt{-1}$, as usual. It suffices to find the product of the constant terms of the polynomials \[(x + i)^4 + a(x+ i)^3 + b(x+i)^2 + c(x+i) + d\]and \[(x - i)^4 + a(x - i)^3 + b(x-i)^2 + c(x-i) + d.\] Calculating these yields the product \[(1 + ai - b - ci + d)(1 - ai - b + ci + d)\]\[= (1 - (b - d))^2 + (a - c)^2\] Since $b - d \geq 5$, we plug in $5$ as the value of $b - d$ and take $a = c$ to yield a minimum value of $4^2 = \boxed{16}$. To show that such a polynomial with real roots and the condition $b - d = 5$ and $a = c$ exists, we present the construction $(x - 1)^4$. The given product evaluates to be $16$ in this case, $b - d = 5$, and $a = c$ as desired.

I wonder which problem came earlier, this or this? Is $2014 > 2024$ or not? This only further verifies this blog post . Construction: The answer is $\boxed{16}$, achieved by $$P(x)=(x+1)^4=x^4+4x^3+6x^2+4x+1\implies (x_1^2+1)(x_2^2+1)(x_3^2+1)(x_4^2+1)=2^4=16.$$We verify that $b-d=6-1\ge 5$. Now, we prove that this is minimal. Proof of Minimality: Notice that $(x_1^2+1)(x_2^2+1)(x_3^2+1)(x_4^2+1)$ can be written as $$(x_1^2+1)(x_2^2+1)(x_3^2+1)(x_4^2+1)=(x_1-i)(x_1+i)(x_2-i)(x_2+i)(x_3-i)(x_3+i)(x_4-i)(x_4+i)$$$$=((i-x_1)(i-x_2)(i-x_3)(i-x_4))\cdot ((-i-x_1)(-i-x_2)(-i-x_3)(-i-x_4))$$$$=P(i)P(-i)$$$$=(1-ai-b+ci+d)(1+ai-b-ci+d)$$$$=(1-b+d)^2-(ai-ci)^2$$$$=(1-(b-d))^2+(a-c)^2$$$$\ge 16+0=16,$$which finishes. JuanOrtiz wrote: This is a truly horrible problem Only when you use Vieta's