We firstly write $d=\frac{ac}{b}$ so we have $\frac{a}{b}+\frac{b}{a}+\frac{b}{c}+\frac{c}{b}=4$ and we want to show that $\frac{a}{c}+\frac{c}{a}+\frac{b^2}{ac}+\frac{ac}{b^2}$ or equivalently $(\frac{a}{b}+\frac{b}{a})(\frac{c}{b}+\frac{b}{c}) \leqslant -12$. Set $\frac{a}{b}+\frac{b}{a}=s$ and $\frac{c}{b}+\frac{b}{c}=t$. We have $s+t=4$ and we want $st \leqslant -12$. We have $t=4-s$ and so we need $s(4-s) \leqslant -12 \Rightarrow s^2-4s-12 \geqslant 0$. If $s \leqslant -2$, this holds (indeed, set $s=-2+k$ and we have $s^2-4s-12=(k-2)^2-4(k-2)-12=k^2-8k-16=(k-4)^2 \geqslant 0$. So ,assume that $s \geqslant -2$. Similarly, assume $t \geqslant -2$. We have then $\frac{a}{b}+\frac{b}{a} \geqslant -2 \Rightarrow \frac{(a+b)^2}{ab}>0$, so $ab>0$ and $bc>0$, so $a,b,c,$ are of the same sign, so $4=\frac{a}{b}+\frac{b}{a}+\frac{b}{c}+\frac{c}{b} \geqslant 2+2=4$, so $a=b=c$, contradiction, since $a \neq b \neq c \neq a$. So, at least one of $s,t$ is $\leqslant -2$ and the stated inequality holds. The equality e.g. when $(a,b,c,d)=(2\sqrt{2}-3,3-2\sqrt{2},1,-1)$ (and this shows that $-12$ is the maximum).

Is this Pan American Math Olympiad?

African^

Dear Sir, you forgot to wrote pairwise distinct, but you use it in solution, so I do similarly. Also it should be show that $-12$ is minimal Notation like above. Observe that $ab>0 \implies a^2+b^2\ge 2ab\iff s\ge 2$ and $ab<0\implies a^2+b^2\ge -2ab\iff s\le -2$. Analogously $t\ge 2$ or $t\le -2$ $s(4-s) \leqslant -12\iff (s+2)(s-6)\ge 0$ If $s\le -2$, then the inequality is true. Now assume $s\ge 2$. If $s=2\iff a=b$ so contradiction, as $a,b,c,d$ are all pairwise distinct. Hence $4-t=s>2\iff 2>t$ and that's equivalent to $-2\ge t$ as we proved above. So $-st-12=-(4-t)t-12=(t+2)(t-6)\ge 0$

Motivation for equality is $s=-2$, so $a=-b$. Then $c=-d$ and $t=4-s=6$. Of course one can do $t=-2$ and so on.

Yes, you are right WolfusA, I will corect. Also, I had written a case where equality is achived, and that shows that it is indeed the maximum.

As an alternative solution, we can verify that the quantities $\frac{a}{b}$, $\frac{b}{c}$, $\frac{c}{d}$, and $\frac{d}{a}$ are distinct [not quite; see below], that none of them are equal to $1$, and (by using Vieta's relations) that they are the roots of the polynomial $$ x^4 - 4x^3 + (A + 2)x^2 - 4x + 1 $$where $$ A = \frac{a}{c} + \frac{b}{d} + \frac{c}{a} + \frac{d}{b}. $$ We have that $$ x^4 - 4x^3 + (A + 2)x^2 - 4x + 1 = 0 \iff \left(x + \frac{1}{x}\right)^2 - 4\left(x + \frac{1}{x} \right) + A = 0 $$ This has four distinct solutions, so two of the solutions correspond to the equation $$ x + \frac{1}{x} = 2 + \sqrt{4 - A} $$and two of the solutions correspond to $$ x + \frac{1}{x} = 2 - \sqrt{4 - A}. $$ Since none of the roots is equal to $1$, we have that $x + \frac{1}{x} \neq 2$, and so $\sqrt{4 - A}$ is strictly positive. Thus we have that $2 - \sqrt{4 - A} < 2$. For the solutions corresponding to $$ x + \frac{1}{x} = 2 - \sqrt{4 - A}, $$we must have that $x$ is negative since if $x$ were positive then we would have that $x + \frac{1}{x} \geq 2$. For negative $x$ we have that $x + \frac{1}{x} \leq -2$, and so we have that $\sqrt{4 - A} \geq 4$, which gives us that $A \leq -12$. For equality to occur, we must have that $\sqrt{4 - A} = 4$, and so two solutions of the equation occur when $$ x + \frac{1}{x} = -2 \iff x = -1 $$and two solutions of the equation occur when $$ x + \frac{1}{x} = 2 + \sqrt{4 - A} = 6 \iff x = 3 \pm \sqrt{8}. $$ We can thus take $\frac{a}{b} = \frac{c}{d} = -1$ and $\frac{b}{c} = \frac{a}{d} = 3 + \sqrt{8}$. One possibility is then $(a, b, c, d) = (3 + \sqrt{8}, -3 - \sqrt{8}, -1, 1)$. edit: We don't have that $\frac{a}{b}$, etc... are distinct, but we do have that $\frac{a}{b} \neq \frac{b}{c}$, and $\frac{c}{d} \neq \frac{d}{a}$, which is enough. We use the "fact" that the roots are distinct to conclude that two roots of the polynomial correspond to $$ x + \frac{1}{x} = 2 + \sqrt{4 - A} $$and that two roots correspond to $$ x + \frac{1}{x} = 2 - \sqrt{4 - A}. $$(In other words, we need to show that the negative solution does in fact occur) These equations have the property that $x$ is a root if and only if $\frac{1}{x}$ is. Now we know that $\frac{a}{b} = \frac{d}{c}$, so $\frac{a}{b}$ and $\frac{c}{d}$ are roots of one of the two equations above. The number $\frac{b}{c}$, which is not equal to either of these two, is a root of the other quadratic.

Orestis_Lignos wrote: We have $t=4-s$ and so we need $s(4-s) \leqslant -12 \Rightarrow s^2-4s-12 \geqslant 0$. If $s \leqslant -2$, this holds (indeed, set $s=-2+k$ and we have $s^2-4s-12=(k-2)^2-4(k-2)-12=k^2-8k-16\color{red}{=}(k-4)^2 \geqslant 0$.

Orestis_Lignos wrote: Let $a,b,c,d$ be nonzero pairwise distinct real numbers, such that $\frac{a}{b}+\frac{b}{c}+\frac{c}{d}+\frac{d}{a}=4$ and $ac=bd$. Show that $\frac{a}{c}+\frac{b}{d}+\frac{c}{a}+\frac{d}{b} \leqslant -12$ and show that $-12$ is the best constant. It was here: https://artofproblemsolving.com/community/c6h466959

Ghoshadi wrote: Orestis_Lignos wrote: We have $t=4-s$ and so we need $s(4-s) \leqslant -12 \Rightarrow s^2-4s-12 \geqslant 0$. If $s \leqslant -2$, this holds (indeed, set $s=-2+k$ and we have $s^2-4s-12=(k-2)^2-4(k-2)-12=k^2-8k-16\color{red}{=}(k-4)^2 \geqslant 0$. This can be easily corrected. Let $f(s)=s^2-4s-12=(s+2)(s-6)$. So,$ f(s)\ge0\Leftrightarrow s\ge6$ or $s\le-2$

Notice that using conditions : $(a+b+c+d)^2= \sum a^2 +12ac $ and prove that $ac<0 $(simple ). then you are done !

Ghoshadi wrote: Orestis_Lignos wrote: We have $t=4-s$ and so we need $s(4-s) \leqslant -12 \Rightarrow s^2-4s-12 \geqslant 0$. If $s \leqslant -2$, this holds (indeed, set $s=-2+k$ and we have $s^2-4s-12=(k-2)^2-4(k-2)-12=k^2-8k-16\color{red}{=}(k-4)^2 \geqslant 0$. Oh, sorry, you are right, hopefully it can be easily corrected, thanks to hansu.

Same problem with here.

lminsl wrote: Same problem with here.