April wrote: Let $ a$, $ b$, $ c$, $ d$ be positive real numbers such that $ abcd = 1$ and $ a + b + c + d > \dfrac{a}{b} + \dfrac{b}{c} + \dfrac{c}{d} + \dfrac{d}{a}$. Prove that \[ a + b + c + d < \dfrac{b}{a} + \dfrac{c}{b} + \dfrac{d}{c} + \dfrac{a}{d}\] Very Easy ! We know that: $ (a + b + c + d).(ab + bc + cd + da) > (\frac {a}{b} + \frac {b}{c} + \frac {c}{d} + \frac {d}{a}).(ab + bc + cd + da) \geq (a + b + c + d)^2$ It follows that $ ab + bc + cd + da > a + b + c + d$ It suffices to prove that: $ \frac {b}{a} + \frac {c}{b} + \frac {d}{c} + \frac {a}{d} \geq ab + bc + cd + da$ But: $ (\frac {b}{a} + \frac {c}{d}) + (\frac {c}{b} + \frac {d}{a}) + (\frac {d}{c} + \frac {a}{b}) + (\frac {a}{d} + \frac {b}{c}) \geq 2.(ab + bc + cd + da) > (ab + bc + cd + da) + (a + b + c + d) > (ab + bc + cd + da) + \frac {a}{b} + \frac {b}{c} + \frac {c}{d} + \frac {d}{a}$ and the result follows immediately.

April wrote: Let $ a$, $ b$, $ c$, $ d$ be positive real numbers such that $ abcd = 1$ and $ a + b + c + d > \dfrac{a}{b} + \dfrac{b}{c} + \dfrac{c}{d} + \dfrac{d}{a}$. Prove that \[ a + b + c + d < \dfrac{b}{a} + \dfrac{c}{b} + \dfrac{d}{c} + \dfrac{a}{d}\] It was also a problem in Russia Winter TST 2008. Russia_Man posted it but received no replies. http://www.mathlinks.ro/viewtopic.php?search_id=1658696324&t=250987. rachid's solution is very nice.

By Cauchy Schwarz Inequality, we have that \[ a + b + c + d > \frac {a}{b} + \frac {b}{c} + \frac {c}{d} + \frac {d}{a} = \frac {a^2}{ab} + \frac {b^2}{bc} + \frac {c^2}{cd} + \frac {d^2}{da} \ge \frac {(a + b + c + d)^2}{ab + bc + cd + da}\] so we have that \[ ab + bc + cd + da > a + b + c + d.\] Now, by Cauchy Schwarz Inequality, we also have that \[ \left(\frac {a}{b} + \frac {b}{c} + \frac {c}{d} + \frac {d}{a}\right)\left(\frac {d}{c} + \frac {a}{d} + \frac {b}{a} + \frac {c}{b}\right) \ge \left(\sqrt {\frac {ad}{bc}} + \sqrt {\frac {ab}{cd}} + \sqrt {\frac {bc}{da}} + \sqrt {\frac {cd}{ab}}\right)^2 = (ad + ab + bc + cd)^2 > (a + b + c + d)^2\] and since $ a + b + c + d > \frac {a}{b} + \frac {b}{c} + \frac {c}{d} + \frac {d}{a}$, then we must have $ a + b + c + d < \frac {b}{a} + \frac {c}{b} + \frac {d}{c} + \frac {a}{d}.$ QED

By AM-GM, $ 4a = 4\sqrt [4]{\frac {a}{d}\frac {a}{b}\frac {a}{b}\frac {b}{c}}\leq\frac {a}{d} + 2\frac {a}{b} + \frac {b}{c}$, and similarly for its cyclic permutations. Adding all of them up, \[ 4(a + b + c + d)\leq3\left(\frac {a}{b} + \frac {b}{c} + \frac {c}{d} + \frac {d}{a}\right) + \frac {b}{a} + \frac {c}{b} + \frac {d}{c} + \frac {a}{d} < 3(a + b + c + d) + \frac {b}{a} + \frac {c}{b} + \frac {d}{c} + \frac {a}{d}.\]

April wrote: Let $ a$, $ b$, $ c$, $ d$ be positive real numbers such that $ abcd = 1$ and $ a + b + c + d > \dfrac{a}{b} + \dfrac{b}{c} + \dfrac{c}{d} + \dfrac{d}{a}$. Prove that \[ a + b + c + d < \dfrac{b}{a} + \dfrac{c}{b} + \dfrac{d}{c} + \dfrac{a}{d}\] Proposed by Slovakia $ \frac{a}{b}+\frac{b}{a} \ge a+b \iff (a-b)^2(a+b) \ge 0$, and so on. So $ \dfrac{a}{b} + \dfrac{b}{c} + \dfrac{c}{d} + \dfrac{d}{a} + \dfrac{b}{a} + \dfrac{c}{b} + \dfrac{d}{c} + \dfrac{a}{d} \ge 2(a+b+c+d)$ from which the problem follows.

Mathias_DK wrote: April wrote: Let $ a$, $ b$, $ c$, $ d$ be positive real numbers such that $ abcd = 1$ and $ a + b + c + d > \dfrac{a}{b} + \dfrac{b}{c} + \dfrac{c}{d} + \dfrac{d}{a}$. Prove that \[ a + b + c + d < \dfrac{b}{a} + \dfrac{c}{b} + \dfrac{d}{c} + \dfrac{a}{d}\] Proposed by Slovakia $ \frac {a}{b} + \frac {b}{a} \ge a + b \iff (a - b)^2(a + b) \ge 0$, It 's wrong.Try $ a=1;b=2$ The solution of daniel 73 very nice

the official solution?

quykhtn-qa1 wrote: Mathias_DK wrote: April wrote: Let $ a$, $ b$, $ c$, $ d$ be positive real numbers such that $ abcd = 1$ and $ a + b + c + d > \dfrac{a}{b} + \dfrac{b}{c} + \dfrac{c}{d} + \dfrac{d}{a}$. Prove that \[ a + b + c + d < \dfrac{b}{a} + \dfrac{c}{b} + \dfrac{d}{c} + \dfrac{a}{d}\] Proposed by Slovakia $ \frac {a}{b} + \frac {b}{a} \ge a + b \iff (a - b)^2(a + b) \ge 0$, It 's wrong.Try $ a = 1;b = 2$ The solution of daniel 73 very nice Oh yes, of course i'm wrong (The idea is the same though)

$(\frac{a}{b}+\frac{b}{c}+\frac{c}{d}+\frac{d}{a})+(\frac{b}{a}+\frac{c}{b}+\frac{d}{c}+\frac{a}{d}) =(a+c)(\frac{1}{b}+\frac{1}{d})+(b+d)(\frac{1}{a}+\frac{1}{c}) =(a+c)(b+d)(\frac{1}{ac}+\frac{1}{bd}) =(a+c)(b+d)(ac+bd)=(a+b+c+d)(\frac{1}{\frac{1}{a+c}+\frac{1}{b+d}})(ac+bd) \ge (a+b+c+d)(\frac{1}{\frac{1}{2\sqrt{ac}}+\frac{1}{2\sqrt{bd}}})(ac+bd) \ge 2(a+b+c+d)$.

WLOG $ac\ge1$. Suppose for the sake of contradiction that $a+b+c+\frac{1}{abc}>\frac{a}{b}+\frac{b}{c}+abc^2+\frac{1}{a^2bc}$ and $a+b+c+\frac{1}{abc}\ge\frac{b}{a}+\frac{c}{b}+\frac{1}{abc^2}+a^2bc$. Expanding, we get that \[(ac^3-c+1)(ab)^2-ac(a+c)(ab)+(a^3c-a+1)<0\]and \[(a^3c-a+1)(bc)^2-ac(a+c)(bc)+(ac^3-c+1)\le0.\]Since $ac\ge1$, $ac^3-c+1\ge c^2-c+1>0$ and $a^3c-a+1\ge a^2-a+1>0$. Consider \[f(x)=(ac^3-c+1)x^2-ac(a+c)x+(a^3c-a+1)=(ac^3-c+1)(x-r)(x-s),\]which must have real roots $r<s$ since $\lim_{x\to\infty}f(x)=\infty$ and $f(ab),f(1/bc)\le0$. Thus $r<ab<s$ and $r\le \frac{1}{bc}\le s$, whence \[r^2 < \frac{a}{c} < s^2\implies f(\sqrt{a/c})<0,\]i.e. letting $u=a+c$ and $v=ac\ge1$ and noting that $u\ge2\sqrt{v}$, \[2v>u(v^2-v^{3/2}+1)\ge2\sqrt{v}(v^2-v^{3/2}+1)\implies (v^{3/2}-1)(v^{1/2}-1)<0,\]a clear contradiction.

April wrote: Let $ a$, $ b$, $ c$, $ d$ be positive real numbers such that $ abcd = 1$ and $ a + b + c + d > \dfrac{a}{b} + \dfrac{b}{c} + \dfrac{c}{d} + \dfrac{d}{a}$. Prove that \[ a + b + c + d < \dfrac{b}{a} + \dfrac{c}{b} + \dfrac{d}{c} + \dfrac{a}{d}\] Proposed by Pavel Novotný, Slovakia Suppose, for the sake of contradiction, that \[{a + b + c + d \ge \frac{b}{a} + \frac{c}{b} + \frac{d}{c} + \frac{a}{d}. \quad \text(1)}.\] Then, by multiplying the given inequality by $(1)^3$, it follows that \[\left(a + b + c + d\right)^4 > \left(\frac{a}{b} + \frac{b}{c} + \frac{c}{d} + \frac{d}{a}\right)\left(\frac{b}{a} + \frac{c}{b} + \frac{d}{c} + \frac{a}{d}\right)^3. \quad \text{(2)}\] However, this is false, since, by Holder's Inequality, we have that \begin{align*} \left(\frac{a}{b} + \frac{b}{c} + \frac{c}{d} + \frac{d}{a}\right)\left(\frac{b}{a} + \frac{c}{b} + \frac{d}{c} + \frac{a}{d}\right)^3 \\ &= \left(\frac{a}{b} + \frac{b}{c} + \frac{c}{d} + \frac{d}{a}\right)\left(\frac{d}{c} + \frac{a}{d} + \frac{b}{a} + \frac{c}{b}\right)\left(\frac{a}{d} + \frac{b}{a} + \frac{c}{b} + \frac{d}{c}\right)\left(\frac{a}{d} + \frac{b}{a} + \frac{c}{b} + \frac{d}{c}\right) \\ &\ge \left(\sqrt[4]{\frac{a^3d}{bcd^2}} + \sqrt[4]{\frac{b^3a}{cda^2}} + \sqrt[4]{\frac{c^3b}{dab^2}} + \sqrt[4]{\frac{d^3c}{abc^2}}\right)^4 \\ &= \left(a + b + c + d\right)^4. \end{align*} Therefore, we conclude that $(2)$ was false, and hence our assumption was flawed. Thus, we see that $\;$ $a + b + c + d < \frac{b}{a} + \frac{c}{b} + \frac{d}{c} + \frac{a}{d}$ $\;$ as desired. $\blacksquare$

Isn't this just holder: $(\frac{b}{a} +...+\frac{a}{d})^3\cdot(\frac{a}{b}+...)(a+b+c+d) \geq (a+b+c+d)^6$ Which is by easy manipulation :$(\frac{b}{a}+...)>a+b+c+d$ PS. I wrote this on phone so thatsbwhy I was lazy about fractions

full solution:http://www.artofproblemsolving.com/community/c42708h1085977_imo_2008_a5

Assume not : $ a+b+c+d \geq \frac{b}{a}+\frac{c}{b}+\frac{d}{c}+\frac{a}{d} $ $ \Rightarrow a+b+c+d>\sqrt{(\frac{b}{a}+\frac{c}{b}+\frac{d}{c}+\frac{a}{d})(\frac{c}{d}+\frac{d}{a}+\frac{a}{b}+\frac{b}{c})} \geq (a+c)(b+d) $ On the other hands : $ (a+b+c+d)(ab+bc+cd+da) \geq (ab+bc+cd+da)(\frac{b}{a}+\frac{c}{b}+\frac{d}{c}+\frac{a}{d}) \geq (a+b+c+d)^2 $ $ \Rightarrow ab+bc+cd+da \geq a+b+c+d \rightarrow\leftarrow $ done!

Solved with william122 Use $\sum$ to denote cyclic sums. The problem follows from the following chain of inequalities: $$\sqrt{\sum \frac{a}{b}\sum \frac{d}{c}} \ge \sum \sqrt{\frac{ad}{bc}} = \sum ad \ge \frac{(\sum a)^2}{\sum \frac{d}{a}} > \sum a.$$

Define the following quantities := $$S_1 = \frac ab +\frac bc +\frac cd +\frac da$$$$S_2 = \frac ba +\frac cb +\frac dc +\frac ad$$$$W =a+b+c+d$$. We claim that $3S_1+S_2 \geq 4W$ . This definitely finishes as then $S_2 \geq W + \underbrace{3(W-S_1)}_{>0} >W$ To prove the inequality we note that$$2 \cdot \frac ab + \frac bc + \frac ad \overset{\text{AM-GM}}{\geq} 4\sqrt [4]{\frac {a}{d}\cdot \frac {a^2}{b^2} \cdot \frac {b}{c}} \geq 4a$$. Cyclically summing up we are done . $\blacksquare$

Posting for storage. Solved with eisirrational, goodbear, and nukelauncher. First note $\sum a > \sum a/b \geq (a+b+c+d)^2/(ab+bc+cd+da)$ or $a+b+c+d < ab+bc+cd+da$. Then since \[(\sum a/b)(\sum d/c) \geq (ad+ba+cb+dc)^2 > (a+b+c+d)^2,\]the two factors on the LHS cannot both be $\leq a+b+c+d$.

The idea is to show \[(a+b+c+d)^2\le (\tfrac ab+\tfrac bc+\tfrac cd+\tfrac da)(\tfrac ba+\tfrac cb+\tfrac dc+\tfrac ad).\]Expand \[(a+b+c+d)^2=\frac{(a+b+c+d)^2}{\sqrt{abcd}}=\sum_{\mbox{cyc}}\sqrt{\frac{a^3}{bcd}}+\frac{1}{2}\sum_{\mbox{sym}}\sqrt{\frac{ab}{cd}}.\]Expand \[(\tfrac ab+\tfrac bc+\tfrac cd+\tfrac da)(\tfrac ba+\tfrac cb+\tfrac dc+\tfrac ad)=4+\sum_{\mbox{cyc}} (\tfrac{a^2}{bd}+\tfrac{bd}{a^2})+\sum_{\mbox{cyc}}\frac{ab}{cd}.\]Write \[1+\frac{ab}{cd}\ge 2\sqrt{\frac{ab}{cd}}\]and symmetric results. Then it suffices to show \[\sum_{\mbox{cyc}} (\tfrac{a^2}{bd}+\tfrac{bd}{a^2})\ge \sum_{\mbox{cyc}}\sqrt{\frac{a^3}{bcd}}+2\sqrt{\frac{ac}{bd}}+2\sqrt{\frac{bd}{ac}}.\]Observe that \[\sum_{\mbox{cyc}}\sqrt{\frac{a^3}{bcd}}+2\sqrt{\frac{ac}{bd}}+2\sqrt{\frac{bd}{ac}}\le \frac{1}{2}\sum_{\mbox{cyc}}(\tfrac{a^2}{bd}+\tfrac{a}{c})+2\sqrt{\frac{ac}{bd}}+2\sqrt{\frac{bd}{ac}}.\]Additionally, observe the AM-GM powered results of \begin{align*} \sum_{\mbox{cyc}}\left(\frac{a^2}{bd}+\frac{bd}{c^2}\right)&\ge 2\sum_{\mbox{cyc}}\frac{a}{c}\\ \sum_{\mbox{cyc}}\frac{bd}{a^2}&\ge 2\sqrt{\frac{bd}{ac}}+2\sqrt{\frac{ac}{bd}}\\ \sum_{\mbox{cyc}}\frac{a^2}{bd}&\ge 2\sqrt{\frac{bd}{ac}}+2\sqrt{\frac{ac}{bd}}.\\ \end{align*}Sum the first inequality multiplied by $\tfrac 14$, the second inequality multiplied by $\tfrac 34$, and the third inequality multiplied by $\tfrac 14$ to conclude the result.

It is given to us that $$a+b+c+d> \frac{a}{b}+\frac{b}{c}+\frac{c}{d}+\frac{d}{a}.$$Assume for the sake of contradiction that $$a+b+c+d\geq \frac{b}{a}+\frac{c}{b}+\frac{d}{c}+\frac{a}{d}.$$Note that by our assumption$,$ $$(a+b+c+d)^2\bigg(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}+\frac{1}{d}\bigg)>\bigg(\frac{a}{b}+\frac{b}{c}+\frac{c}{d}+\frac{d}{a}\bigg)\bigg(\frac{a}{d}+\frac{b}{a}+\frac{c}{b}+\frac{d}{c}\bigg)\bigg(\frac{1}{c}+\frac{1}{d}+\frac{1}{a}+\frac{1}{b}\bigg)=:P.$$ But by $\text{Hölder's inequality},$ $$P\geq \bigg(\sqrt[3]{\frac{a^2}{bdc}}+\sqrt[3]{\frac{b^2}{cad}}+\sqrt[3]{\frac{c^2}{dba}}+\sqrt[3]{\frac{d^2}{acb}}\bigg)^3=(a+b+c+d)^3,$$so it follows that $$\fbox{$\frac{1}{a}+\frac{1}{b}+\frac{1}{c}+\frac{1}{d}> a+b+c+d.$}$$ This time notice that $$(a+b+c+d)^3>(a+b+c+d)\bigg(\frac{b}{c}+\frac{c}{d}+\frac{d}{a}+\frac{a}{b}\bigg)\bigg(\frac{d}{c}+\frac{a}{d}+\frac{b}{a}+\frac{c}{b}\bigg)=:Q.$$ With the use of $\text{Hölder's inequality},$ $$Q\geq \bigg(\sqrt[3]{\frac{abd}{c^2}}+\sqrt[3]{\frac{bca}{d^2}}+\sqrt[3]{\frac{cdb}{a^2}}+\sqrt[3]{\frac{dac}{b^2}}\bigg)^3=\bigg(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}+\frac{1}{d}\bigg)^3,$$from where it follows that $$\fbox{$a+b+c+d>\frac{1}{a}+\frac{1}{b}+\frac{1}{c}+\frac{1}{d}$},$$which was the desired contradiction$.$ $\blacksquare$

dame dame

By Cauchy-Schwarz Inequality, \[(ad+ab+bc+cd)\left(\frac{a}{d}+\frac{b}{a}+\frac{c}{b}+\frac{d}{c}\right)\geqslant (a+b+c+d)^2\]so $ad+ab+bc+cd\geqslant \frac{(a+b+c+d)^2}{\frac{a}{d}+\frac{b}{a}+\frac{c}{b}+\frac{d}{c}}.$ Then \begin{align*} (a+b+c+d)\left(\frac{b}{a}+\frac{c}{b}+\frac{d}{c}+\frac{a}{d}\right)>& \left(\frac{a}{b}+\frac{b}{c}+\frac{c}{d}+\frac{d}{a}\right)\left(\frac{b}{a}+\frac{c}{b}+\frac{d}{c}+\frac{a}{d}\right) \\ \geqslant& \left(\sqrt{\frac{ad}{bc}}+\sqrt{\frac{ab}{cd}}+\sqrt{\frac{bc}{ad}}+\sqrt{\frac{cd}{ab}}\right)^2 \\ =& (ad+ab+bc+cd)^2 \\ \geqslant& \frac{(a+b+c+d)^4}{\left(\frac{a}{d}+\frac{b}{a}+\frac{c}{b}+\frac{d}{c}\right)^2}. \end{align*}Thus, \[ a + b + c + d < \dfrac{b}{a} + \dfrac{c}{b} + \dfrac{d}{c} + \dfrac{a}{d}. \quad \blacksquare\]

Solved with Max Lu and Ryan Yang. First note by Cauchy-Schwarz that \begin{align*} (a+b+c+d)(ab+bc+cd+da) &>\left(\frac ab+\frac bc+\frac cd+\frac da\right)(ab+bc+cd+da)\\ &\ge(a+b+c+d)^2, \end{align*}implying that \(ab+bc+cd+da>a+b+c+d\). It follows, again by Cauchy-Schwarz, that \begin{align*} (a+b+c+d) \left(\frac ba+\frac cb+\frac dc+\frac ad\right) &> \left(\frac ab+\frac bc+\frac cd+\frac da\right) \left(\frac dc+\frac ad+\frac ba+\frac cb\right)\\ &\ge\left(\sqrt{\frac{da}{bc}}+\sqrt{\frac{ab}{cd}}+\sqrt{\frac{bc}{da}}+\sqrt{\frac{cd}{ab}}\right)^2\\ &=(da+ab+bc+cd)^2 \ge(a+b+c+d)^2, \end{align*}implying \(\frac ba+\frac cb+\frac dc+\frac ad>a+b+c+d\) as desired.

Stephen wrote: $$(\frac{a}{b}+\frac{b}{c}+\frac{c}{d}+\frac{d}{a})+(\frac{b}{a}+\frac{c}{b}+\frac{d}{c}+\frac{a}{d}) =(a+c)(\frac{1}{b}+\frac{1}{d})+(b+d)(\frac{1}{a}+\frac{1}{c}) $$$$=(a+c)(b+d)(\frac{1}{ac}+\frac{1}{bd}) =(a+c)(b+d)(ac+bd)=(a+b+c+d)(\frac{1}{\frac{1}{a+c}+\frac{1}{b+d}})(ac+bd) $$$$\ge (a+b+c+d)(\frac{1}{\frac{1}{2\sqrt{ac}}+\frac{1}{2\sqrt{bd}}})(ac+bd) \ge 2(a+b+c+d)$$ https://artofproblemsolving.com/community/c6h1280948p23908388 Let $a,b,c,d>0.$Prove that $$ \dfrac{a}{b} + \dfrac{b}{c} + \dfrac{c}{d} + \dfrac{d}{a} + \dfrac{b}{a} + \dfrac{c}{b} + \dfrac{d}{c} + \dfrac{a}{d} \geq 2\sqrt{\left( \dfrac{a}{b} + \dfrac{b}{c} + \dfrac{c}{d} + \dfrac{d}{a}\right )\left( \dfrac{b}{a} + \dfrac{c}{b} + \dfrac{d}{c} + \dfrac{a}{d}\right)}$$$$\geq \sqrt{2\left( \dfrac{1}{ac} + \dfrac{1}{bd}\right )}\\\ (a+b+c+d)\geq \frac{2(a+b+c+d)}{\sqrt[4]{abcd}}$$(SXTX (11)2021)

oops Suppose that $a+b+c+d>\frac ab+\frac bc+\frac cd+\frac da$ and $a+b+c+d\geq\frac ba+\frac cb+\frac dc+\frac ad$. Let $a=\frac wx$, $b=\frac xy$, $c=\frac yz$, and $d=\frac zw$. Then, we have \begin{align*} \frac wx+\frac xy+\frac yz+\frac zw&>wy\left(\frac1{x^2}+\frac1{z^2}\right)+xz\left(\frac1{y^2}+\frac1{w^2}\right)\\ \frac wx+\frac xy+\frac yz+\frac zw&\geq\frac{x^2+z^2}{wy}+\frac{w^2+y^2}{xz}, \end{align*}so \begin{align*} w^2yz+x^2zw+y^2wx+z^2xy&>\frac{w^2y^2z}x+\frac{w^2xy^2}z+\frac{wx^2z^2}y+\frac{x^2yz^2}w\\ w^2yz+x^2zw+y^2wx+z^2xy&\geq x^3z+xz^3+w^3y+wy^3. \end{align*} However, by the AM-GM Inequality, we have \begin{align*} &w^2yz+x^2zw+y^2wx+z^2xy\\ >&\frac34(x^3z+xz^3+w^3y+wy^3)+\frac14\left(\frac{w^2y^2z}x+\frac{w^2xy^2}z+\frac{wx^2z^2}y+\frac{x^2yz^2}w\right)\\ =&\left(\frac12w^3y+\frac14z^3x+\frac14\frac{y^2zw^2}x\right)+\left(\frac12x^3z+\frac14w^3y+\frac14\frac{z^2wx^2}y\right)+\left(\frac12y^3w+\frac14x^3z+\frac14\frac{w^2xy^2}z\right)+\left(\frac12z^3x+\frac14y^3w+\frac14\frac{x^2yz^2}w\right)\\ \geq&w^2yz+x^2zw+y^2wx+z^2xy. \end{align*} This is a contradiction. Therefore, it is impossible for $a+b+c+d\geq\frac ba+\frac cb+\frac dc+\frac ad$, so we must have $a+b+c+d<\frac ba+\frac cb+\frac dc+\frac ad$.

Note that \[(a+b+c+d)(ab+bc+cd+da)> \left(\dfrac{a}{b} + \dfrac{b}{c} + \dfrac{c}{d} + \dfrac{d}{a}\right)(ab+bc+cd+da)\ge (a+b+c+d)^2\]by Cauchy-Schwarz. Thus, $ab+bc+cd+da>a+b+c+d.$ It remains to show that $ab+bc+cd+da\le \dfrac{b}{a} + \dfrac{c}{b} + \dfrac{d}{c} + \dfrac{a}{d}.$ Now, note that \[\left(\dfrac{b}{a} + \dfrac{c}{b} + \dfrac{d}{c} + \dfrac{a}{d}\right)\left(\dfrac{a}{b} + \dfrac{b}{c} + \dfrac{c}{d} + \dfrac{d}{a}\right)\ge \left(\sqrt{\frac{bc}{ad}}+\sqrt{\frac{cd}{ab}}+\sqrt{\frac{da}{bc}}+\sqrt{\frac{ab}{cd}}\right)^2\]We have \[\left(\sqrt{\frac{bc}{ad}}+\sqrt{\frac{cd}{ab}}+\sqrt{\frac{da}{bc}}+\sqrt{\frac{ab}{cd}}\right)^2=\left(\frac{bc+da}{\sqrt{abcd}}+\frac{cd+ab}{\sqrt{abcd}}\right)^2=(ab+bc+cd+da)^2>(a+b+c+d)^2.\]The result, thus follows.

By C-S: $$\left(\sum_{\text{cyc}} \frac{a}{b}\right )\left (\sum_{\text{cyc}} \frac{b}{a}\right )\geqslant \left (\sum_{\text{cyc}} \sqrt{\frac{ad}{bc}}\right ) \left (\sum_{\text{cyc}} \sqrt{\frac{ad}{bc}}\right )= \left (\sum_{\text{cyc}} ad\right ) \left (\sum_{\text{cyc}} ad\right )\geqslant \left (\frac{\left (a+b+c+d\right )^2}{\sum ab^{-1}} \right)^2 > \left (\sum_{\text{cyc}} a\right) \left (\sum_{\text{cyc}} a\right)$$

Note the inequality $$\frac{3}{8}\cdot \frac{a}{b}+\frac{3}{8}\cdot \frac{a}{d}+\frac{1}{8}\cdot \frac{b}{c}+\frac{1}{8}\cdot \frac{d}{c} \geq \frac{a}{\sqrt[4]{abcd}}=a$$by weighted AM-GM. Summing cyclically and doubling, it follows that $$\frac{a}{b}+\frac{b}{c}+\frac{c}{d}+\frac{d}{a}+\frac{b}{a}+\frac{c}{b}+\frac{d}{c}+\frac{a}{d}\geq 2(a+b+c+d),$$so if $\frac{a}{b}+\frac{b}{c}+\frac{c}{d}+\frac{d}{a}<a+b+c+d$ we must have $\frac{b}{a}+\frac{c}{b}+\frac{d}{c}+\frac{a}{d}>a+b+c+d$. $\blacksquare$

Got a hint to use weighted AM-GM why did I not think of that i guess im rusty We have $$4a=4\sqrt[4]{\frac{a^4}{abcd}}=4\sqrt[4]{\frac{a^3}{bcd}}\le\frac ad+\frac bc+2\frac ab\implies \sum_{cyc}4a\le\sum_{cyc}\frac ad+2\frac ab+\frac bc=\sum_{cyc}3\frac ab+\frac ba<\sum_{cyc}3a+\frac ba,$$which readily implies the result. $\blacksquare$

Note that \[ \frac{3}{8} \cdot \frac{a}{b} + \frac{3}{8}\cdot \frac{a}{d} + \frac{1}{8}\cdot \frac{d}{c} + \frac{1}{8}\cdot \frac{b}{c} \geq a^{\frac{3}{4}}b^{-\frac{1}{4}}c^{-\frac{1}{4}}d^{-\frac{1}{4}} = a \]So summing cyclically yields that \[ \frac{1}{2}\left( \frac{b}{a} + \frac{c}{b} + \frac{d}{c} + \frac{a}{d}\right) + \frac{1}{2}\left( \frac{a}{b} + \frac{b}{c} + \frac{c}{d} + \frac{d}{a}\right) \geq a + b + c + d \]Using the given condition implies that $\frac{b}{a} + \frac{c}{b} + \frac{d}{c} + \frac{a}{d} \geq a + b +c + d$. $\blacksquare$

Generalization 1 Let $a_{1},a_{2},\cdots,a_{n}$ be positive reals such that $\prod{a_1}=1$ and $$\sum_{cyc}{a_{1}}> \sum_{cyc-j}{\dfrac{a_{j}}{a_{j+1}}}$$equalities hold. Then prove that $$\sum_{cyc}{a_1}< \sum_{cyc-j}{\dfrac{a_{j}^2a_{j+1}a_{j+2}}{\prod{a_1}}}$$

Let $S_1 = \dfrac{a}{b} + \dfrac{b}{c} + \dfrac{c}{d} + \dfrac{d}{a}$ and $S_2 = \dfrac{b}{a} + \dfrac{c}{b} + \dfrac{d}{c} + \dfrac{a}{d}$ Note that $S_1 = \frac{ad+bc}{bd}+\frac{ab+cd}{ac}$ and $S_2 = \frac{bc+ad}{ac}+\frac{cd+ab}{bd}$. So, we get that $$S_1+S_2 = (b+d)(a+c)\left(\frac{1}{ac}+\frac{1}{bd}\right) \geq 2(b+d)(a+c)$$ Now, by Cauchy Schwarz inequality, $$\left(\sqrt{\frac{a}{c}}+\sqrt{\frac{b^2}{ac}}+\sqrt{\frac{c}{a}}+\sqrt{\frac{d^2}{ac}}\right)^2 \leq \left(\frac{a}{b}+\frac{b}{c}+\frac{c}{d}+\frac{d}{a}\right)\left(\frac{b}{c}+\frac{b}{a}+\frac{d}{a}+\frac{d}{c}\right) = S_1\frac{(b+d)(a+c)}{ac}$$multiplying by $ac$ we get that $$(a+b+c+d)^2 \leq (b+d)(a+c)S_1$$Similarly, $(a+b+c+d)^2 \leq (b+d)(a+c)S_2$, and taking the mean of the $2$ inequalities we get that $$(a+b+c+d)^2 \leq (b+d)(a+c)\left(\frac{S_1+S_2}{2}\right) \leq \left(\frac{S_1+S_2}{2}\right)^2$$ implying that if $S_1 < a+b+c+d$ then $S_2 > a+b+c+d$

If we show $2(a+b+c+d)\le \frac ab+\frac bc+\frac cd+\frac da+\frac ba+\frac cb+\frac dc+\frac ad$ we would be done. This rearranges to $(a+c)\left(2-\frac1b-\frac1d\right)+(b+d)\left(2-\frac1a-\frac1c\right)\le 0.$ Dividing by $(a+c)(b+d)$ turns this into $\frac2{b+d}+\frac2{a+c}\le\frac1{ac}+\frac1{bd}.$ However notice that replacing $a,c$ by $\sqrt{ac},\sqrt{ac}$ increases the left side so assuming $a=c,b=d$ makes it $\frac1{a^2}+\frac1{b^2}\ge\frac1a+\frac1b$ for positive reals $ab=1.$ Squaring and homogenizing this becomes $\frac b{a^3}+\frac a{b^3}\ge\frac1{b^2}+\frac1{a^2}.$ This following by summing AM-GM $\frac34\frac b{a^3}+\frac14\frac a{b^3}\ge\sqrt[4]{\left(\frac b{a^3}\right)^3\frac a{b^3}}=\frac1{b^2},\frac14\frac b{a^3}+\frac34\frac a{b^3}\ge\sqrt[4]{\frac b{a^3}\left(\frac a{b^3}\right)^3}=\frac1{a^2}.$

Our condition tells us \[a+b+c+d > \frac{a^2}{ab} + \frac{b^2}{bc} + \frac{c^2}{cd} + \frac{d^2}{da} \ge \frac{(a+b+c+d)^2}{ab+bc+cd+da},\] so $ab+bc+cd+da > a+b+c+d$. Cauchy now gives us \[\left(\frac ab + \frac bc + \frac cd + \frac da\right) \left(\frac dc + \frac ad + \frac ba + \frac cb\right) \ge\left(\frac{da+ab+bc+cd}{\sqrt{abcd}}\right)^2 > (a+b+c+d)^2,\] from which we have the desired by dividing out the given condition. $\blacksquare$

The key claim is the following: Claim: We have \[ \left( \sum_{cyc} \frac ab \right)^{\frac 32} \left( \sum_{cyc} \frac ad \right)^{\frac 32} \left( \sum_{cyc} \frac bc \right)^{\frac 12} \left( \sum_{cyc} \frac dc \right)^{\frac 12} \ge \left( \sum_{cyc} a^{\frac 34} b^{-\frac 14} c^{-\frac 14} d^{-\frac 14} \right)^4 \]Proof. This is due to Holder's inequality. $\blacksquare$ Our claim is equivalent to \[ \left( \sum_{cyc} \frac ab \right)^2 \left( \sum_{cyc} \frac ad \right)^2 \ge \left( \sum_{cyc} a \right)^4 \]where we used $(abcd)^{\frac 14} = 1$. But then, this is equivalent to \[ \left( \sum_{cyc} \frac ab \right) \left( \sum_{cyc} \frac ad \right) \ge (a+b+c+d)^2 \]In particular, then \[ \sum_{cyc} \frac ad \ge \frac{(a+b+c+d)^2}{\sum_{cyc} \frac ab} > a+b+c+d \]as needed. $\blacksquare$ Remark [Motivation]. I recently saw a few problems that involved using Holder to throw terms together, possibly from the same sum, hoping for the product to be a term on the right hand side. I guess the same strategy works here.

Posting this because I found it funny (Top 10 reasons to post fr) \[ a+b+c+d>\frac{a}{b}+\frac{b}{c}+\frac{c}{d}+\frac{d}{a}=\frac{a^2}{ab}+\frac{b^2}{bc}+\frac{c^2}{cd}+\frac{d^2}{ad} \overset{T2}{\ge} \frac{(a+b+c+d)^2}{ab+bc+cd+da} \implies ab+bc+cd+da>a+b+c+d \]\[ \left(\frac{a}{b}+\frac{b}{c}+\frac{c}{d}+\frac{d}{a} \right) \left(\frac{d}{c}+\frac{a}{d}+\frac{b}{a}+\frac{c}{b} \right) \overset{C-S}{\ge} \left(\sqrt{\frac{ad}{bc}}+\sqrt{\frac{ab}{dc}}+\sqrt{\frac{bc}{ad}}+\sqrt{\frac{cd}{ab}} \right)^2=(ad+ba+bc+cd)^2>(a+b+c+d)^2 \]This clearly finishes due to the initial condition, thus we are done .