$ \text{LHS}=(ab+1)(bc+1)(cd+1)(da+1)$ By AM-GM: $ (ab+1)(cd+1)=2+ab+cd\ge 2+2\sqrt{abcd}=4$ Simirlary: $ (bc+1)(da+1)\ge 4$ Hence: $ \text{LHS}\ge 16$ Equality occurs if $ a=c=t,b=d=\frac{1}{t},t\in [\frac{1}{2},2]$ Please check this solution, i am not wrong ????

What is MEMO, Yimin Ge? It seems to be a very interesting competition, with many nice inequalities? For this problem, we have to find the max (max, not min, quang) of \[ (ab+1)(bc+1)(cd+1)(da+1)=(2+ab+cd)(2+ad+bc)=(2+ab+\frac{1}{ab})(2+ad+\frac{1}{ad}).\] WLOG, assume that $ ac\le 1$ and $ b\ge d$ then $ bd=\frac{1}{ac}$ and \[ b,d\in [\frac{1}{2ac}, 2]\] It is easy to show that our expression attains the maximum if $ \{b,d\}=\{2,\frac{1}{2ac}\}$, namely \[ (2+ab+\frac{1}{ab})(2+ad+\frac{1}{ad})\le (2+2a+\frac{1}{2a})(2+\frac{1}{2c}+2c).\] Let $ m=2a,n=2c$ then $ m,n\in[1,mn]$ and $ mn\le 4$. We have \[ (2+m+\frac{1}{m})(2+n+\frac{1}{n})\le (2+1+\frac{1}{1})(2+mn+\frac{1}{mn})=4(2+mn+\frac{1}{mn})\le 4(2+4+\frac{1}{4})=27.\] The maximum attains for $ a=c=1, b=2,d=\frac{1}{2}.$ Just I found this solution when I stay facing with the computer and have no pen to draft. I hope the solution is true.

Using the substitution $ a =\frac{x}{y}, b =\frac{y}{z}, c =\frac{z}{t}, d =\frac{t}{x}$, we have to find the maximum value of $ E = (\frac{x}{z}+\frac{z}{x}+2)(\frac{y}{t}+\frac{t}{y}+2)$ for $ 1\le x,y,z,t\le 2$. Since $ \frac{x}{z}+\frac{z}{x}\le\frac{5}{2}$ and $ \frac{y}{t}+\frac{t}{y}\le\frac{5}{2}$, the maximum value is $ \frac{81}{4}$. As Hungkhtn shown above, the maximum value is attainded for $ (a,b,c,d)=(1,2,1,\frac {1}{2})$.

erm obiously something is wrong but not sure what. $ f(2,2,\frac{1}{2},\frac{1}{2}) = 25$ try inserting in original inequality. and $ 25 >\frac{81}{4}$. Vasc how do u know that $ 1\leq x,y,z,t\leq2$? i was under the impression that if u chose a substitution like that $ x,y,z,t$ the only thing u can say is that $ 2min(x,y,z,t)\geq max(x,y,z,t)$ hungkhtn why can u assume $ ac\leq 1$ and $ b\geq d$? i think u could assume that only if inequality was symethrical not cyclical. btw MEMO = Middle European Mathematical Olimpiad its the first time for this competition i think. anyway my solution goes like this: $ (ab+1)(bc+1)(cd+1)(da+1) = 4+\frac{a}{c}+\frac{c}{a}+\frac{b}{d}+\frac{d}{b}+2(ab+bc+ca+da)$ - now the max of $ \frac{a}{c}+\frac{c}{a}$ is when $ a$ and $ c$ are as far away from eash other as posible. example $ a = 2, c =\frac{1}{2}$. same goes for $ \frac{b}{d}+\frac{d}{b}$ $ b = 2, d =\frac{1}{2}$ - now $ ab+bc+ca+da = (a+c)(b+d) <\left(\frac{a+b+c+d}{2}\right)^{2}$ equality holds if $ a+c = b+d$ - max of $ a+b+c+d$ if $ abcd = k$ is when $ a,b,c,d$ are as much appart as thay can. so conviniently one of them is $ 2$ and one of them is $ \frac{1}{2}$. Its obivous now that max of $ a+b+c+d$ occurs when 2 are $ 2$ and two are $ \frac{1}{2}$. combining all three points we see that max of inequality occurs at $ f(a,b,c,d) = f(2,2,\frac{1}{2},\frac{1}{2}) = 25$

Well, I can assume that $ ac\le 1$ and $ b\ge d$ actually. First, we can assume that $ ac\le 1$ (of course), since $ ac$ and $ bd$ have the same role. Then, between $ b$ and $ d$, we can assume $ b\ge d$ of course (since the role of $ b$ and $ d$ are the same. Anyway, the maximum is actually taken when $ a=1,b=2,c=1,d=\frac{1}{2}$. I am sure that.

if $ a=1,b=2,c=1,d=\frac{1}{2}$ than: $ f(1,2,1,\frac{1}{2})=\left(1+\frac{1}{2}\right)\left(2+\frac{1}{1}\right)\left(1+\frac{1}{\frac{1}{2}}\right)\left(\frac{1}{2}+\frac{1}{1}\right)=(\frac{3}{2})(3)(3)(\frac{3}{2})=\frac{81}{4}$ while if $ a=2,b=2,c=\frac{1}{2},d=\frac{1}{2}$ we have: $ f(2,2,\frac{1}{2},\frac{1}{2})=\left(2+\frac{1}{2}\right)\left(2+\frac{1}{\frac{1}{2}}\right)\left(\frac{1}{2}+\frac{1}{\frac{1}{2}}\right)\left(\frac{1}{2}+\frac{1}{2}\right)=(\frac{5}{2})(4)(\frac{5}{2})(1)=25$ since $ 25>\frac{81}{4}$ than: $ f(2,2,\frac{1}{2},\frac{1}{2})>f(1,2,1,\frac{1}{2})$ so maximum can't be at $ 1,2,1,\frac{1}{2}$ Is this so or where did i make a mistake? But i agree now that there's nothing wrong with the assumption $ ac\leq 1$ and $ b\geq d$ so i don't know who of us and where made a mistake

Jure the frEEEk wrote: Vasc how do u know that $ 1\leq x,y,z,t\leq2$? You are right, Jure the frEEk. This is a sufficient condition, but not necessary. Therefore, my assumption is wrong.

From $ (4a-c)(4c-a)\ge 0$, we get $ a+c\le\frac {5}{2}\sqrt{ac}$ and $ \frac {a}{c}+\frac {c}{a}\le\frac{17}{4}$. Similarly, $ b+d\le\frac {5}{2}\sqrt{bd}$ and $ \frac {b}{d}+\frac {d}{b}\le\frac{17}{4}$. Then, $ (a+c)(b+d)\le\frac{25}{4}\sqrt{abcd}=\frac{25}{4}$, and hence $ (ab+1)(bc+1)(cd+1)(da+1) = 4+\frac{a}{c}+\frac{c}{a}+\frac{b}{d}+\frac{d}{b}+2(a+c)(b+d)\le$ $ 4+\frac{17}{4}+\frac{17}{4}+2\frac{25}{4}=25$.

Well, it is so funny that I made this mistake and no found. I am sorry, my mistake is \[ (2+m+\frac{1}{m})(2+n+\frac{1}{n})\le (2+1+\frac{1}{1})(2+mn+\frac{1}{mn})=4(2+mn+\frac{1}{mn})\le 4(2+4+\frac{1}{4})=27.\] While, it should be \[ 4(2+4+\frac{1}{4})=25.\] Such a stupid thing! Anyway, 25 is the final solution. The equality holds for $ m=1,n=4$, or $ a=\frac{1}{2},c=2,b=2,d=\frac{1}{2}.$

well lol. this things happen. I was actually searching for ur mistake and also didn't see it

Multiplying $ (a+\frac{1}{b})\cdot(c+\frac{1}{d})$, doing the same with $ (b+\frac{1}{c})\cdot(d+\frac{1}{a})$, and using $ abcd=1$ a bit you can get the following equation: $ (a+\frac{1}{b})\cdot(b+\frac{1}{c})\cdot(c+\frac{1}{d})\cdot(d+\frac{1}{a}) = (\sqrt{\frac{a}{c}}+\sqrt{\frac{c}{a}}+\sqrt{\frac{b}{d}}+\sqrt{\frac{d}{b}})^{2}$ now it's easily seen that maximum is 25, and it is taken for $ (\frac{1}{2},\frac{1}{2}, 2, 2)$ and for $ (2,2,\frac{1}{2},\frac{1}{2})$.

quangpbc wrote: $ \text{LHS} = (ab + 1)(bc + 1)(cd + 1)(da + 1)$ By AM-GM: $ (ab + 1)(cd + 1) = 2 + ab + cd\ge 2 + 2\sqrt {abcd} = 4$ Simirlary: $ (bc + 1)(da + 1)\ge 4$ Hence: $ \text{LHS}\ge 16$ Equality occurs if $ a = c = t,b = d = \frac {1}{t},t\in [\frac {1}{2},2]$ Please check this solution, i am not wrong ???? I think you found the minium of LHS!