Is $\frac{a+b}{c+d}<\frac{a}{c}+\frac{b}{d}?$, for $a,b,c,d>0$? Is $$\frac{a+b}{c+d}<\frac{a}{c}+\frac{b}{d}$$
for $a,b,c,d>0$
If it is true, then can we generalize?
EDIT:typing mistake corrected.
EDIT, WILL JAGY. Apparently the real question is 
Is $$\color{magenta}{\frac{a+b}{c+d}<\frac{a}{c}+\frac{b}{d}}$$
for $a,b,c,d>0,$ where letters on the left hand side and in the numerator stay in the numerator on the right-hand side, and letters on the left hand side and in the denominator stay in the denominator on the right-hand side.
 A: If you consider them as slopes, then $(0,0)$, $(b,a)$, $(d,c)$ and $(b+d,a+c)$ form a parallelogram. So the slope of the line between $(0,0)$ and $(b+d,a+c)$ will be between the slopes of the lines between $(0,0)$ and $(b,a)$ and $(d,c)$. That means that $\frac{a+c}{b+d}$ will be between $\frac{a}{b}$ and $\frac{c}{d}$. Since these two are positive, this means that $$\frac{a+c}{b+d}\leq \max\left(\frac{a}{b},\frac{c}{d}\right)< \frac{a}{b}+\frac{c}{d}$$
It's pretty clear that you can generalize this by induction to:
$$\frac{a_1+\dots+a_n}{b_1+\dots+b_n}\leq \max_i\left(\frac{a_i}{b_i}\right)$$
A: A slightly different approach:
Multiply both sides by (c+d), which we can do without altering the inequality because c and d are positive:
$$ a+b < \frac{a(c+d)}{c} +\frac{b(c+d)}{d}$$
$$ a+b < \frac{ac}{c} +\frac{ad}{c} +\frac{bc}{d} +\frac{bd}{d}$$
$$ a+b < a + \frac{ad}{c} +\frac{bc}{d} +b$$
$$ a+b < a+b +\frac{ad}{c}+\frac{bc}{d}$$
$$ \frac{ad}{c} +\frac{bc}{d} > 0$$
This is clearly always true because both terms must be > 0.
This same basic outline works for a 3 term version of this:
$$\frac{a+b+c}{d+e+f} < \frac{a}{d}+\frac{b}{e} +\frac{c}{f}$$
and will clearly work for any number of terms because after multiplying by the denominator on the left hand side, you will always spit out on the right hand side, exactly the left hand side numerator plus some additional terms which must be positive.
A: Let $a=\lambda c$ and $b=\mu d$, where $\lambda, \mu>0$.
$$\frac {a+b}{c+d}=\frac {\lambda c+\mu d}{c+d}=\frac {\lambda (c+d)+\mu (c+d)-(\lambda d+\mu c)}{c+d}=\lambda + \mu -\frac {\lambda d+\mu c}{c+d}<\lambda + \mu = \frac ac+\frac bd
$$
