Let n≥2,n∈N, a,b,c,d∈N, ab+cd<1 and a+c≤n, find the maximum value of ab+cd for fixed n.
Problem
Source: China TST 1993, problem 2
Tags: quadratics, function, algebra unsolved, algebra, inequalities
18.09.2006 14:25
It is easy. Let s=ab+cd=ad+bcbd. If b=d then s≤1−1b≤1−12n. If b≠d then s≤bd−1bd=1−1bd. From a+c≤n we get b,d≤2n. We have n−1n+1n+1=1−1n(n+1).
16.01.2007 04:05
Rust wrote: From a+c≤n we get b,d≤2n. We have n−1n+1n+1=1−1n(n+1). I don't understand that, can you help me?
16.01.2007 08:19
Let s=ab+cd=ad+bcbd<1. If b=d then s=a+cb, maximum when b=a+c+1, because a+c≤n, s≤1−1n+1. If b≠d then s≤bd−1bd=1−1bd. If b≥n+1,c≥n+1, then s≤1−1n+1. Therefore if s>1−1n+1 one of b,d less than n+1. Let b≤n. If d≤n, then s≤1−1n2. If d>n+2, then s=1−b−ab+cd≤1−b−ab+n−an+2<1−1n(n+1). We have n−1n+1n+1=1−1n(n+1). Therefore it give maximum s.
12.03.2014 14:25
Not true unfortunately. For n=5,for example, we have 2/9+3/4=35/36 > 29/30. The exact answer is a bit complicated. The maximum equals 1-1/t where t(3p)=4p^3+4p^2+3p+1 t(3p+1)=4p^3+8p^2+7p+2 t(3p+2)=4p^3+12p^2+14p+6 So t is not quadratic,but of third power wrt. n
13.03.2014 07:06
can you give detail solution alexheinis?
14.03.2014 14:08
I included the details in a pdf-file, I only had the solution in Dutch so I translated. Bye.
Attachments:
detail.pdf (84kb)
16.03.2014 19:58
Setting b=a+x,d=c+y the problem gets easier, because we have aa+x+cc+y<1⟺xy>ac. We assume a≤c. Keeping π:=xy fixed, we will show that the form aa+x+cc+y is maximal for x=π,y=1. We have aa+π+cc+1≥aa+x+cc+y⟺ay+cx≤a+cπ. Using the convex function f(x)=cx+aπx one sees that the maximum on [1,π] is attained in one of the extremities and, in this case, when x=π. Hence, if xy=π, then the form is at most aa+π+cc+1. This form is maximal when π=ac+1, substitution yields 1−1t, where t:=(ac+a+1)(c+1). We want to maximise t, it is increasing in a,c, hence we may assume a+c=n. With γ:=c+1,σ:=n+1 we find γ(aγ+1) and we consider g(x)=x+x2(σ−x) on [σ2,σ]. Then g is concave on this interval with a maximum in x=σ+√σ2+33. Notice 2σ3<x<2σ+13 and because we consider integers, we split up with regards to n modulo 3. If n=3p+1 we have σ=3p+2 and 2p+1<x<2p+2. We need to plug in x=2p+1,2p+2 and check which x yields the bigger value. Similarly for the other cases modulo 3. We find t(3p)=4p3+4p2+3p+1,t(3p+1)=4p3+8p2+7p+2,t(3p+2)=4p3+12p2+14p+6
19.12.2015 11:34
Let a,b,c,d∈N and ab+cd<1. Prove that ab+cd<1−1(a+c)3.
24.07.2018 06:34
可以用n的代数式表示。