It is easy. Let $s=\frac{a}{b}+\frac{c}{d}=\frac{ad+bc}{bd}$. If b=d then $s\le 1-\frac{1}{b}\le 1-\frac{1}{2n}$. If $b\not =d$ then $s\le \frac{bd-1}{bd}=1-\frac{1}{bd}.$ From $a+c\le n$ we get $b,d\le 2n$. We have $\frac{n-1}{n}+\frac{1}{n+1}=1-\frac{1}{n(n+1)}$.

Rust wrote: From $a+c\le n$ we get $b,d\le 2n$. We have $\frac{n-1}{n}+\frac{1}{n+1}=1-\frac{1}{n(n+1)}$. I don't understand that, can you help me?

Let $s=\frac{a}{b}+\frac{c}{d}=\frac{ad+bc}{bd}<1$. If b=d then $s=\frac{a+c}{b}$, maximum when $b=a+c+1$, because $a+c\le n$, $s\le 1-\frac{1}{n+1}$. If $b\not =d$ then $s\le \frac{bd-1}{bd}=1-\frac{1}{bd}.$ If $b\ge n+1,c\ge n+1$, then $s\le 1-\frac{1}{n+1}$. Therefore if $s>1-\frac{1}{n+1}$ one of $b,d$ less than n+1. Let $b\le n$. If $d\le n$, then $s\le 1-\frac{1}{n^{2}}$. If $d>n+2$, then $s=1-\frac{b-a}{b}+\frac{c}{d}\le 1-\frac{b-a}{b}+\frac{n-a}{n+2}<1-\frac{1}{n(n+1)}$. We have $\frac{n-1}{n}+\frac{1}{n+1}=1-\frac{1}{n(n+1)}$. Therefore it give maximum s.

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

can you give detail solution alexheinis?

I included the details in a pdf-file, I only had the solution in Dutch so I translated. Bye.

Attachments:
detail.pdf (84kb)

Setting $b=a+x,d=c+y$ the problem gets easier, because we have ${a\over {a+x}}+{c\over {c+y}}<1\iff xy>ac$. We assume $a\le c$. Keeping $\pi:=xy$ fixed, we will show that the form ${a\over {a+x}}+{c\over {c+y}}$ is maximal for $x=\pi,y=1$. We have ${a\over {a+\pi}}+{c\over{c+1}}\ge {a\over {a+x}}+{c\over {c+y}}\iff ay+cx\le a+c\pi$. Using the convex function $f(x)=cx+{{a\pi}\over x}$ one sees that the maximum on $[1,\pi]$ is attained in one of the extremities and, in this case, when $x=\pi$. Hence, if $xy=\pi$, then the form is at most ${a\over {a+\pi}}+{c\over {c+1}}$. This form is maximal when $\pi=ac+1$, substitution yields $1-{1\over t}$, 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 $\gamma:=c+1,\sigma:=n+1$ we find $\gamma(a\gamma+1)$ and we consider $g(x)=x+x^2(\sigma-x)$ on $[{\sigma\over 2},\sigma]$. Then $g$ is concave on this interval with a maximum in $x={{\sigma+\sqrt{\sigma^2+3}}\over 3}$. Notice ${{2\sigma}\over 3}<x<{{2\sigma+1}\over 3}$ and because we consider integers, we split up with regards to $n$ modulo 3. If $n=3p+1$ we have $\sigma=3p+2$ 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 $$t(3p)=4p^3+4p^2+3p+1,t(3p+1)=4p^3+8p^2+7p+2,t(3p+2)=4p^3+12p^2+14p+6$$

Let $a,b,c,d \in \mathbb{N}$ and $\frac{a}{b}+\frac{c}{d}<1$. Prove that $$\frac{a}{b}+\frac{c}{d}<1-\frac{1}{(a+c)^3}.$$

可以用n的代数式表示。