Can we say that $\frac{a+b}{c+d} \leq \frac{a}{c} + \frac{b}{d} $

Question

If $a,b,c,d$ are positive real numbers, can we say that $\frac{a+b}{c+d} \leq \frac{a}{c} + \frac{b}{d} $ is always true? If no, can you please give insignts on under which conditions this might be true.

Any references to a similar type of inequalities are also welcome

Thank you,

Answer 1

Solution to new version:

Since $$\frac{a}{c+d} \leq \frac{a}{c}$$ and $$\frac{b}{c+d} \leq \frac{b}{d}$$ we have $$\frac{a+b}{c+d} = \frac{a}{c+d} + \frac{b}{c+d} \leq \frac{a}{c} + \frac{b}{d} $$

Answer 2

This is an answer to first version of the problem: Does following holds $${a+b\over c+d}\leq {a\over b}+{c\over d}$$

No, put $a=c= 1$ and $b=d=3$ and we get $${1+3\over 1+3}\nleq {1\over 3}+{1\over 3}$$

Answer 3

After the latest edit:

Wlog. $\frac ac\le\frac bd$. Then $$\frac ac\le\frac{a+b}{c+d}\le\frac bd<\frac ac+\frac bd$$

Indeed, $\frac ac\le \frac bd$ implies $\frac{bc-ad}{cd}\ge 0$, i.e., $bc-ad\ge 0$. Then $$ \frac bd-\frac{a+b}{c+d}=\frac{b(c+d)-d(a+b)}{d(c+d)}=\frac{bc-ad}{d(c+d)}\ge0$$ and $$ \frac{a+b}{c+d}-\frac ac=\frac{c(a+b)-a(c+d)}{(c+d)a}=\frac{bc-ad}{(c+d)a} \ge 0$$