ab+bc+cd+da=(a+c)(b+d)< or equal to 2*10^1/2. so 2*10^1/2 is the max value. when a+c= b+d

The searched maximum value is $9$. By AM-GM we have \[(a+c)^2+(b+d)^2=a^2+b^2+c^2+d^2+2(ac+bd)\geq 10+4\sqrt{abcd}=18\] so \[(a+c)(b+d)\leq\dfrac{(a+c)^2+(b+d)^2}{2}=9\] Equal holds when if and only if $ac=bd, a+c=b+d$, which implies that $(a,c)\sim(1,2)$ and $(b,d)\sim(1,2)$.

gxggs wrote: The searched maximum value is $9$. By AM-GM we have \[(a+c)(b+d)\leq\dfrac{(a+c)^2+(b+d)^2}{2}=9\] This is not true, you have the $\geq$ sign instead of $=$, which is not easy to fix.

That is really a silly mistake......

94337sk wrote: ab+bc+cd+da=(a+c)(b+d)< or equal to 2*10^1/2. so 2*10^1/2 is the max value. when a+c= b+d $ab+bc+cd+da=(a+c)(b+d)\le 2\sqrt{10}$ when $ a+c= b+d$$?$

This problem is really interesting. I am sure that \[(a+c)(b+d)\leq \sqrt{82}\] Equal holds when $a=c=\sqrt{\frac{5}{2}}$ and $b=\sqrt{\frac{5}{2}+\frac{3\sqrt{41}}{10}}, d=\sqrt{\frac{5}{2}-\frac{3\sqrt{41}}{10}}$. My proof for this problem is very simple and only a few lines.

It is not difficult..Since a2*+b2*+c2*+d2* grater then or equal to ab+bc+cd+ad .And so max is 10.

Vo Quoc Ba Can's Proof http://voquocbacan.blogspot.in/2013/04/inequality-from-czech-polish-slovak.html?spref=fb

Sayan wrote: Vo Quoc Ba Can's Proof http://voquocbacan.blogspot.in/2013/04/inequality-from-czech-polish-slovak.html?spref=fb Thank Sayan. China can not see.

I don't see the proof of Vo Quoc Ba Can. I post my proof here: Denote that $x=ab+bc+cd+da=(a+c)(b+d)$, by AM-GM we have \begin{align*} x=&(a+c)(b+d)=\sqrt{(a^2+c^2+2ac)(b^2+d^2+2bd)}\\ =&\sqrt{(a^2+c^2)(b^2+d^2)+2ac(b^2+d^2)+2bd(a^2+c^2)+4abcd}\\ \leq & \sqrt{\left(\frac{a^2+b^2+c^2+d^2 }{2}\right)^2+2(ab+cd)(ad+bc)+16}\\ \leq &\sqrt{25+2\cdot\left(\frac{ab+cd+ad+bc}{2}\right)^2+16}\\ =&\sqrt{41+\frac{1}{2}x^2} \end{align*} Solving this inequality gives us that $x\leq \sqrt{82}$. I omit finding equal cases here.

gxggs wrote: I post my proof here: Denote that $x=ab+bc+cd+da=(a+c)(b+d)$, by AM-GM we have \begin{align*} x=&(a+c)(b+d)=\sqrt{(a^2+c^2+2ac)(b^2+d^2+2bd)}\\ =&\sqrt{(a^2+c^2)(b^2+d^2)+2ac(b^2+d^2)+2bd(a^2+c^2)+4abcd}\\ \leq & \sqrt{\left(\frac{a^2+b^2+c^2+d^2 }{2}\right)^2+2(ab+cd)(ad+bc)+16}\\ \leq &\sqrt{25+2\cdot\left(\frac{ab+cd+ad+bc}{2}\right)^2+16}\\ =&\sqrt{41+\frac{1}{2}x^2} \end{align*} Solving this inequality gives us that $x\leq \sqrt{82}$. I omit finding equal cases here. Beautiful. Thanks.

Let $a,b,c,d>0, ab+bc+cd+da = \sqrt{82} $ and $ a^2 + b^2 + c^2 + d^2 =10.$ Prove that$$ 4\leq a b c d \leq \frac{41}{8}$$