Solution (Storage, essentially same as above): We claim that the minimum is $8$. Note that $\frac{a}{b}+\frac{b}{c}+\frac{c}{d}+\frac{d}{a} \ge 2(\sqrt{\frac{ac}{bd}} + \sqrt{\frac{bd}{ac}}) = \frac{2(ac+bd)}{\sqrt{abcd}} = \frac{2(a+c)(b+d)}{\sqrt{abcd}} \ge 2*\frac{2\sqrt{ac}*2 \sqrt{bd}}{\sqrt{abcd}} = 8$, where the last inequality is true by AM-GM. Note that equality occurs when $a=c$ and $b=d$. $\square$

Let $F(a,b,c,d)=\frac{a}{b}+\frac{b}{c}+\frac{c}{d}+\frac{d}{a}$. We have $\frac{a}{b}+\frac{c}{d} \geq 2\frac{\sqrt{ac}}{\sqrt{bd}}$ and $\frac{b}{c}+\frac{d}{a} \geq 2 \frac{\sqrt{bd}}{\sqrt{ac}}$. Hence, $$F(a,b,c,d)\geq F(\sqrt{ac}, \sqrt{bd}, \sqrt{ac}, \sqrt{bd})=\frac{2(ac+bd)}{\sqrt{abcd}}=\frac{2(a+c)(b+d)}{\sqrt{abcd}} \geq 8,$$where the second to last equality follows from the condition, and the last inequality follows from $a+c \geq 2\sqrt{ac}$ and $b+d \geq 2 \sqrt{bd}$. Equality is achieved if and only if $\frac{a}{b}=\frac{c}{d}$, $\frac{b}{c}=\frac{d}{a}$, $a=c$ and $b=d$. All of these conditions are actually equivalent to $a=c$ and $b=d$. If $a=c$ and $b=d$, then the condition from the problem is $4ab=a^2+b^2$, which has a solution in positive reals. Therefore, $8$ is the smallest possible value.

how is this an A3, should of not been on the shortlist

Well uh I'll just nuke this like everyone above, I guess We use AM-GM. Verify that $$\frac ab + \frac bc + \frac cd + \frac da \geq 2\left(\sqrt{\frac{ac}{bd}} + \sqrt{\frac{bd}{ca}}\right) = 2\cdot\frac{ac+bd}{\sqrt{abcd}} = 2\cdot \frac{(a+c)(b+d)}{\sqrt{abcd}} \geq 2\cdot\frac{4\sqrt{abcd}}{\sqrt{abcd}} = 8,$$finishing the problem. Equality holds when $a=c$ and $b=d$, which give real $a, b, c, d$ respectively. $\square$

We claim that $\frac ab+\frac bc+\frac cd+\frac da\ge8$, equality achieved when $(a,b,c,d)=\left(2+\sqrt3,1,2+\sqrt3,1\right)$. By AM-GM: $$\frac ab+\frac bc+\frac cd+\frac da\ge2\sqrt{\frac{ac}{bd}}+2\sqrt{\frac{bd}{ac}},$$so it suffices to show that: $$\sqrt{\frac{ac}{bd}}+\sqrt{\frac{bd}{ac}}\ge4$$$$\Leftrightarrow\frac{ac}{bd}+\frac{bd}{ac}\ge14.$$But we have $$\frac{ac}{bd}+\frac{bd}{ac}=\frac{(ac+bd)^2}{abcd}-2=\frac{(a+c)^2(b+d)^2}{abcd}-2,$$and $$\frac{(a+c)^2(b+d)^2}{abcd}-2\ge\frac{(2\sqrt{ac})^2(2\sqrt{bd})^2}{abcd}-2=14,$$implying the result.

I claim that the min value is 8, achievable with $a=c=2+\sqrt{3}$ and $b=d=1$. The pre-condition becomes \[(a+c)(b+d)=8+4\sqrt{3}=ac+bd\]And, \[\frac{a}{b}+\frac{b}{c}+\frac{c}{d}+\frac{d}{a}=2(2+\sqrt{3})+2(\frac{1}{2+\sqrt{3}})=8\]We now show that this is optimal, homogenize so that $abcd=1$, then \[\frac{a}{b}+\frac{b}{c}+\frac{c}{d}+\frac{d}{a}=(\frac{a}{b}+\frac{c}{d})+(\frac{b}{c}+\frac{d}{a})= (\frac{ad+bc}{bd})+\frac{ab+cd}{ac}\]Multiplying by 1, \[=(ac(ad+bc)+bd(ab+cd)\]By AM-GM, \[\geq ac\cdot 2\sqrt{adbc}+bd\cdot 2 \sqrt{abcd}= 2(ab+cd)\]Using pre-condition, \[=2(a+c)(b+d)\geq 2\cdot (2\sqrt{ac})(2\sqrt{bd})\]Thus, $\frac{a}{b}+\frac{b}{c}+\frac{c}{d}+\frac{d}{a}\geq 8$, so we are done.

The answer is $8$, achieved when $(a,b,c,d)=(1,1,2+\sqrt{3},2+\sqrt{3})$. \newline\newline We now prove the bound. Indeed, $$ac+bd=(a+c)(b+d)\geq 4\sqrt{acbd}$$Hence $$\frac{a}{b}+\frac{b}{c}+\frac{c}{d}+\frac{d}{a}\geq 2\frac{\sqrt{ac}}{\sqrt{bd}}+2\frac{\sqrt{bd}}{\sqrt{ac}}=8$$

Posting for storage $$\frac{a}{b}+\frac{b}{c}+\frac{c}{d}+\frac{d}{a} \geq 2 \left(\sqrt{\frac{ac}{bd}}+\sqrt{\frac{bd}{ac}} \right)=2 \left(\frac{ac+bd}{\sqrt{abcd}} \right)=2 \left(\frac{(a+c)(b+d)}{\sqrt{abcd}} \right) \geq 8$$ Equality achieved at $(1,1,2+\sqrt3,2+\sqrt3)$ @below ig this is the most natural approach.

Everyone has the same soluion lol.

AM-GM separately on $\frac ab$ and $\frac cd$ and $\frac bc$ and $\frac da$ gives $$\frac ab + \frac cd \geq 2\sqrt{\frac{ac}{bd}}, \frac bc + \frac da \geq 2\sqrt{\frac{bd}{ca}},$$and adding we get $$\frac ab + \frac bc + \frac cd + \frac da \geq 2\frac{(ac+bd)}{\sqrt{abcd}}=2\frac{(a+c)(b+d)}{\sqrt{abcd}}\geq 2 \cdot 4 \frac{\sqrt{abcd}}{\sqrt{abcd}} = 8.$$Equality at $a=c$ and $b=d$.

Let $abcd=1$ WLOG. We rewrite the expression we seek to minimize as \[\frac{ac}{bc}+\frac{bd}{cd}+\frac{ac}{ad}+\frac{bd}{ab} = ac\cdot \left(\frac{1}{bc}+bc\right)+bd\cdot \left(\frac{1}{cd}+cd\right)\ge 2(ac+bd)=2(ab+bc+cd+da)\ge 8.\]We have equality by taking $a=1,b=2+\sqrt{3},c=1,d=2+\sqrt{3}$.

Here is the same solution presented in the other way. My first motivation is $ac$ and $bd$ cannot be "too close" since $ac+bd=(a+c)(b+d)\geq 4\sqrt{ac\cdot bd}\rightarrow \frac{ac}{bd}\not\in(7-4\sqrt{3},7+4\sqrt{3})$, so we fix $ac=m^2$ and $bd=n^2$. Substituting $a=mk,c=\frac{m}{k},b=nl,d=\frac{n}{l}$. The condition is reduced to $\left(k+\frac{1}{k}\right)\left(l+\frac{1}{l}\right)=\frac{m}{n}+\frac{n}{m}$, and $S$ become $$\frac{m}{n}\left(\frac{k}{l}+\frac{l}{k}\right)+\frac{n}{m}\left(kl+\frac{1}{kl}\right).$$Now, we can bound $\frac{k}{l}+\frac{l}{k}\geq 2$ and $kl+\frac{1}{kl}\geq 2$. Then, we get the conclusion: $$S\geq 2\left(\frac{m}{n}+\frac{n}{m}\right)=2\left(k+\frac{1}{k}\right)\left(l+\frac{1}{l}\right)\geq 2\cdot 2\cdot 2=8.$$The equality hold when $k=l=1$ and $m,n$ be any positive real such that $\frac{m}{n}+\frac{n}{m}=4$.

Problem proposed by AoPS user arqady.

Solution that i found in the TST: So the answer is 8 Proving bound: a/b +b/c +c/d +d/a > 8 × 1=8 ×(ac+bd)/(a+c)(b+d) Now after briefly expanding , we need to prove abcd(a²+b²+c²+d²)+2abcd(ac+bd)+a³d²c+b³a²d+c³b²a+d³c²b+a²c²d²+b²a²d²+ +c²a²b²+d²b²c²>8abcd(ac+bd) By AM>GM we know abcd(a²+c²)+abcd(b²+d²)>2abcd(ac+bd) (a³d²c+c³b²a)+(d³c²b+b³a²d)>2abcd(ac+bd) (a²c²d²+c²a²b²)+(b²a²d²+d²b²c²)> 2abcd(ac+bd) Summing these inequalities will give desired bound

Same as everyone else - posting for storage

EDIT: Thanks for clarifying @1below @6below

math90 wrote: Proposed my Michael Rosenberg (arqady). Maestro

Let $a,b,c,d$ are positive real numbers satisfying $(a+c)(b+d)=ac+\frac{2}{3}bd.$ Prove that $$\frac{a}{b}+\frac{b}{c}+\frac{c}{d}+\frac{d}{a}\geq \frac{30-\sqrt{30}}{3}$$Let $a,b,c,d$ are positive real numbers satisfying $(a+c)(b+d)=ac+\frac{3}{4}bd.$ Prove that $$\frac{a}{b}+\frac{b}{c}+\frac{c}{d}+\frac{d}{a}\geq \frac{28-\sqrt{13}}{3}$$

Leo.Euler wrote: We claim the minimal value of the expression to be $8$. We have \[ \frac{a}{b}+\frac{b}{c}+\frac{c}{d}+\frac{d}{a} = \bigg(\frac{a}{b}+\frac{c}{d}\bigg)+\bigg(\frac{b}{c}+\frac{d}{a}\bigg) \ge 2\bigg(\sqrt{\frac{ac}{bd}}+\sqrt{\frac{bd}{ac}}\bigg)=2\bigg(\frac{ac+bd}{\sqrt{abcd}}\bigg)=2\bigg(\frac{a+c}{\sqrt{ac}}\bigg)\bigg(\frac{b+d}{\sqrt{bd}}\bigg) \ge 2(2)(2)=8 \]for all positive $a, b, c, d$. For equality, use $a=c=2+\sqrt{3}$ and $b=d=2-\sqrt{3}$, which fixes the minimal value to be $8$, as claimed. Nice. And correct

Let $ a,b,c,d>0 $ and $\frac{a}{b}+\frac{b}{c}+\frac{c}{d}+\frac{d}{a}=8.$ Prove that $$(a+c)(b+d)\geq ac+bd$$$$(a+c)(b+d)\geq \frac{4(6-\sqrt 3)}{33}(ac+2bd)$$$$(a+c)(b+d)\geq \frac{2(4-\sqrt 3)}{13}(ac+3bd)$$

The answer is $8$ with equality with $a=c=2-\sqrt3$, $b=d=1$; keeping in mind the equality case we AM-GM as follows: $$\frac ab+\frac bc+\frac cd+\frac da=\left(\frac ab+\frac cd\right)+\left(\frac bc+\frac da\right)\geq2\sqrt{\frac{ac}{bd}}+2\sqrt{\frac{bd}{ac}}=2\cdot\frac{ac+bd}{\sqrt{abcd}}=2\cdot\frac{a+c}{\sqrt{ac}}\cdot\frac{b+d}{\sqrt{bd}}\geq2\times2\times2=8.$$ Seems like almost everyone has the same solution...

sqing wrote: Let $ a,b,c,d>0 $ and $\frac{a}{b}+\frac{b}{c}+\frac{c}{d}+\frac{d}{a}=8.$ Prove that $$(a+c)(b+d)\geq ac+bd$$ Solution.For sake of a contradiction, assume $$(a+c)(b+d)<ac+bd.$$Thus we get $$4\sqrt{abcd}\le(a+c)(b+d)<ac+bd,$$from which it follows that \begin{align*} \frac{a}{b}+\frac{b}{c}+\frac{c}{d}+\frac{d}{a} \ge&2\sqrt{\frac{ac}{bd}}+2 \sqrt{\frac{bd}{ac}}\\ =& \frac{2(ac+bd)}{\sqrt{abcd}}\\ >&\frac{2\cdot4\sqrt{abcd}}{\sqrt{abcd}}=8, \end{align*}contradicting the original condition $ \frac{a}{b}+\frac{b}{c}+\frac{c}{d}+\frac{d}{a} =8$. $\blacksquare$

Let $ a,b,c,d>0 .$ Prove that $$\frac{a}{b}+\frac{b}{c}+\frac{c}{d}+\frac{d}{a}\geq \frac{8(ac+bd)}{(a+c)(b+d)}$$(Jichen)

$\frac{a}{b} + \frac{b}{c} + \frac{c}{d} + \frac{d}{a} \geq 2\sqrt\frac{ac}{bd} + 2\sqrt\frac{bd}{ac} = 2\frac{ac+bd}{\sqrt{abcd}} \geq 2 \frac{ac+bd}{(\frac{ab + bc + cd + ad}{4})} = 8.$ "=" holds when $a=c$, $b=d$, and $(\frac{a}{b})^2 = \pm \sqrt3+2$.

Let $E=\frac ab+\frac bc+\frac cd+\frac da$. By the look of the problem, it looks like equality might hold when $a=c$ and $b=d$. Then, we have $a^2+b^2=4ab$ and hence $\frac ab\in\left\{2-\sqrt3,2+\sqrt3\right\}$ which means $E=8$. We will show that this is the answer. Note that $\frac ab+\frac cd\geq2\sqrt{\frac{ac}{bd}}$ and $\frac bc+\frac da\geq2\sqrt{\frac{bd}{ac}}$. Adding the two, we get \[\frac E2\geq\sqrt{\frac{ac}{bd}}+\sqrt{\frac{bd}{ac}}=\frac{ac+bd}{\sqrt{abcd}}=\frac{(a+c)(b+d)}{\sqrt{abcd}}\geq\frac{2\sqrt{ac}\cdot 2\sqrt{bd}}{\sqrt{abcd}}=4\Longrightarrow E\geq\boxed{8}\]So, minimum value is $8$. $\blacksquare$

Our answer is $\boxed{8}$, achievable with $(2-\sqrt 3, 1, 2-\sqrt 3, 1)$. Notice our given expression can be rewritten as \begin{align*} \left(\frac ab + \frac cd\right) + \left(\frac bc + \frac da\right) &\ge 2 \left(\sqrt{\frac{ac}{bd}} + \sqrt{\frac{bd}{ac}}\right) = 2 \left(\frac{ac+bd}{\sqrt{abcd}}\right) \\ &= \frac{2(a+c)(b+d)}{\sqrt{abcd}} \ge \frac{2 \cdot 2\sqrt{ac} \cdot 2\sqrt{bd}}{\sqrt{abcd}} = 8. \quad \blacksquare \end{align*}

Note that the problem is invariant under the transformation $(a,b,c,d)\mapsto (ta,tb,tc,td)$. So we can assume wlog that $abcd=1$. Let $ac=x, a+c=y$ then by the conditions we have $bd=\frac 1x, b+d=\frac 1y \left(x+\frac 1x\right )$. We remark that, by AM-GM, we have $$\frac{(x^2+1)^2}{4x}\geq y^2\geq 4x.$$In particular we have $x^2+1\geq 4x$. This implies that $x\leq 2-\sqrt 3$ or $x\geq 2+\sqrt 3$. Solving for $a,b,c,d$ and computing a bit we find $$\frac ab+\frac bc+\frac cd+\frac da=\frac{(x^2+1)^2}{2x^2}\pm \frac 12 \left (\frac{1}{x^2}-1\right)\sqrt{\frac{(y^2-4x)((x^2+1)^2-4xy^2)}{y^2}}.$$Call this $f_{+}(x,y)$ if we take the plus sign, and $f_{-}(x,y)$ if we take the minus. Consider the function $g(t)=\frac{(t-4x)((x^2+1)^2-4xt)}{t}$ for $t\in \left [4x,\frac{(x^2+1)^2}{4x}\right ]$. It's clear that the minimum of this function is $0$, attained when $t=4x$. And we can show by calculus that the maximum is $(x^2+1-4x)^2$, attained when $t=1+x^2$. Remembering this, we can now minimize $f_{-},f_{+}$: When $x\leq 2-\sqrt 3$, $$f_{+}(x,y)\geq f_{+}(x,\sqrt{4x})=\frac{(x^2+1)^2}{2x^2}\geq \frac{16x^2}{2x^2}=8$$and $$f_{-}(x,y)\geq f(x,\sqrt{x^2+1})=\frac{(x^2+1)^2}{2x^2}-\frac 12 \left (\frac{1}{x^2}-1\right)\left(1+x^2-4x\right)$$and we can show, by calculus or otherwise, that the minimum of this is also $8$. The case when $x\geq 2+\sqrt 3$ is similar and the answer is $8$ as well.

Equality holds when $a=c$ and $b=d$

No way it's an A3! Same as the solutions above, posting for storage.

The answer is $8$, which is attained at $a=c$ and $b=d$ (one can verify that this actually yields real solutions for $a$, $b$, $c$, and $d$). Now, we will show that $8$ is minimal by applying AM-GM twice. Note that $$\bigg(\frac{a}{b}+\frac{c}{d}\bigg)+\bigg(\frac{b}{c}+\frac{d}{a}\bigg)$$$$\ge 2\bigg(\sqrt{\frac{ac}{bd}}+\sqrt{\frac{bd}{ac}}\bigg)$$$$=\frac{2(ac+bd)}{\sqrt{abcd}}$$$$=\frac{2(a+c)(b+d)}{\sqrt{abcd}}$$$$\ge 2\cdot \frac{2\sqrt{ac}\cdot 2\sqrt{bd}}{\sqrt{abcd}}$$$$=8.$$

We have $$\frac{a}{b}+\frac{b}{c}+\frac{c}{d}+\frac{d}{a} \ge 2\left(\sqrt{\frac{ac}{bd}} + \sqrt{\frac{bd}{ac}}\right) = 2\frac{ab + bc + cd + da}{\sqrt{abcd}} \ge \boxed{8},$$where both the inequalities are AM-GM. Equality holds for say $(a, b, c, d) = (1, 2 + \sqrt{3}, 1, 2 + \sqrt{3})$, so we're done.