Strengthen inequality

For convenience, let us to think that $a\le c$ and $t=d-c\in\mathbb{N}$.

1. Not hard to see that if $\frac{a}{b}+\frac{c}{c+t}<1$ then \begin{equation} \frac{a}{b}+\frac{c}{c+t}\le\frac{a}{\left\lfloor \frac{ac}{t}\right\rfloor+a+1} + \frac{c}{c+t} < 1. \end{equation}

2. Now, let us fix only $a\le c$ and let $t\ge1$ is variable. We can show that the sum $$\frac{a}{\left\lfloor \frac{ac}{t}\right\rfloor+a+1}+\frac{c}{c+t}$$ has the maximal value for $t=1$, or by other words $$\frac{t}{c+t}-\frac{a}{\left\lfloor \frac{ac}{t}\right\rfloor+a+1}\geq\frac{1}{c+1}-\frac{a}{a(c+1)+1}=\frac{1}{(c+1)(a(c+1)+1)}.$$

2.a Let $t\le c^2$. Note, that $\left\lfloor \frac{ac}{t}\right\rfloor = \frac{ac}{t} - \frac{ac(\mathrm{mod\ }t)}{t}\ge \frac{ac}{t} - \frac{t-1}{t}$, so $$\frac{t}{c+t}-\frac{a}{\left\lfloor \frac{ac}{t}\right\rfloor+a+1}\ge\frac{t}{c+t}-\frac{at}{a(t+c)+1} = \frac{t}{(c+t)(a(t+c)+1)},$$ and we have to prove that $$\frac{t}{(c+t)(a(t+c)+1)}\ge\frac{1}{(c+1)(a(c+1)+1)}$$ or $$(t-1)\left(t-\frac{ac^2+c}{a}\right)\le0$$ or $$t < c^2+\frac{c}{a}.$$ Case 2.a is done.

2.b Now, let $t \ge c^2+1$. Since $a\le c$, $\left\lfloor\frac{ac}{t}\right\rfloor=0$, therefore we have to prove that $$\frac{t}{c+t}-\frac{a}{a+1}\geq\frac{1}{c+1}-\frac{a}{a(c+1)+1 }.$$ In view of $t\ge c^2+1$, it suffices to prove $$\frac{1}{1+a}-\frac{1}{1+c}\ge\frac{1}{1+c+\frac{1}{c}}-\frac{1}{1+c+\frac{1}{a}}.$$ It is easy to see that for the function $f(x)=\frac1x$ the following property is true:

• If $y\ge x>0$ then for all $\delta > 0$ : $f(x)-f(x+\delta)\ge f(y)-f(y+\delta)$.

From the last statement and from $\frac{1}{c}+c-a\geq \frac{1}{a}$ we can conclude $$\frac{1}{1+a}-\frac{1}{1+c}\ge\frac{1}{1+c+\frac{1}{c}}-\frac{1}{1+c+\frac{1}{c}+(c-a)}\ge \frac{1}{1+c+\frac{1}{c}}-\frac{1}{1+c+\frac{1}{a}}$$

Now 2. is done.

3. From 1. and 2. : $$\frac{a}{b}+\frac{c}{c+t}\le \frac{a}{\left\lfloor\frac{ac}{t}\right\rfloor+a+1}+\frac{c}{c+t}\le \frac{a}{a(c+1)+1}+\frac{c}{c+1}=1-\frac{1}{(c+1)(a(c+1)+1)}\le 1-\frac{1}{(a+c)^3} < 1-\frac{1}{n^3}$$

I hope that more laconic solvation exists:)