$a,b,c,d$ are fixed positive real numbers. Find the maximum value of the function $f: \mathbb{R^{+}}_{0} \rightarrow \mathbb{R}$ $f(x)=\frac{a+bx}{b+cx}+\frac{b+cx}{c+dx}+\frac{c+dx}{d+ax}+\frac{d+ax}{a+bx}, x \geq 0$
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Tags: inequalities, algebra
Circumcircle
17.04.2019 10:36
\bump ....
mihaig
03.05.2019 13:59
The authors are: Leonard Giugiuc,Dan Stefan Marinescu from Romania, Hideyuki Ohtsuka Japan, under the direct coordination of professor Valmir B.Krasniqi, from the University of Prishtina,Kosovo.
mihaig
03.05.2019 23:23
Any echoes? Of course we have the solution,otherwise it couldn't be accepted .It's a real problem,with solution.
mihaig
04.05.2019 16:06
Leartia wrote: $a,b,c,d$ are fixed positive real numbers. Find the maximum value of the function $f: \mathbb{R^{+}}_{0} \rightarrow \mathbb{R}$ $f(x)=\frac{a+bx}{b+cx}+\frac{b+cx}{c+dx}+\frac{c+dx}{d+ax}+\frac{d+ax}{a+bx}, x \geq 0$ A very important hint,directly from the author. DO NOT use derivatives. No chance...
Pathological
04.05.2019 19:42
Let $a_1, a_2, a_3, a_4$ denote $\frac{a}{b}, \frac{b}{c}, \frac{c}{d}, \frac{d}{a},$ respectively. We claim that the maximal value of $f$ on the nonnegative reals is $a_1 + a_2 + a_3 + a_4.$ This is attained at $x = 0.$ Now, we will show it is maximal. Observe that:
$$\frac{a+bx}{b+cx} = \frac{ba_1 + bx}{b + \frac{bx}{a_2}} = \frac{a_1 + x}{1 + \frac{x}{a_2}} = \frac{a_1a_2 + xa_2}{a_2 + x} = a_1 + \frac{x(a_2-a_1)}{a_2 + x}.$$
Therefore, to show the desired inequality, we just need to show that $\sum_{cyc} \frac{a_2 - a_1}{a_2 + x} \le 0.$ Indeed, observe that:
$$\sum_{cyc} \frac{a_2 - a_1}{a_2 + x} = 4 - \sum_{cyc} \frac{a_1+x}{a_2 + x} \le 4 - 4 = 0,$$
where we used the fact that $\sum_{cyc} \frac{a_1 + x}{a_2 + x} \ge 4$ by AM-GM. We are now done.
$\square$
Note: Interestingly, I think we never used the condition that $a_1a_2a_3a_4 = 1.$
mihaig
04.05.2019 20:04
Bravo! It's exactly my question regarding $uvwt=1$ .This I call the miracle of Analysis or so called absorbing property of $0$ .
dangerousliri
04.05.2019 23:47
Pathological wrote:
Let $a_1, a_2, a_3, a_4$ denote $\frac{a}{b}, \frac{b}{c}, \frac{c}{d}, \frac{d}{a},$ respectively. We claim that the maximal value of $f$ on the nonnegative reals is $a_1 + a_2 + a_3 + a_4.$ This is attained at $x = 0.$ Now, we will show it is maximal. Observe that:
$$\frac{a+bx}{b+cx} = \frac{ba_1 + bx}{b + \frac{bx}{a_2}} = \frac{a_1 + x}{1 + \frac{x}{a_2}} = \frac{a_1a_2 + xa_2}{a_2 + x} = a_1 + \frac{x(a_2-a_1)}{a_2 + x}.$$
Therefore, to show the desired inequality, we just need to show that $\sum_{cyc} \frac{a_2 - a_1}{a_2 + x} \le 0.$ Indeed, observe that:
$$\sum_{cyc} \frac{a_2 - a_1}{a_2 + x} = 4 - \sum_{cyc} \frac{a_1+x}{a_2 + x} \le 4 - 4 = 0,$$
where we used the fact that $\sum_{cyc} \frac{a_1 + x}{a_2 + x} \ge 4$ by AM-GM. We are now done.
$\square$
Note: Interestingly, I think we never used the condition that $a_1a_2a_3a_4 = 1.$
This solution is almost the same as official but I have to admit that your solution is much better, also the official one is nice.
mihaig
05.05.2019 00:00
dangerousliri wrote: This solution is almost the same as official but I have to admit that your solution is much better, also the official one is nice. I'm glad this remark comes from oane of the top Kosovo mathematician and trainer. Because if I would told that,no one would believed me.
dangerousliri
05.05.2019 00:07
What I like about this problem is that it is hard but with a simple solution which is rare.
mihaig
05.05.2019 00:18
Yes.And I'm rare too. So rare,that some time I don't find myself. The good joke was done. You and Valmir are extraordinary persons and matematicians. My collaboration with him (through him with you indirectly) was first due to http://www.mathproblems-ks.org/ I saw there some of the works of both of you and I was impressed .Bravo!