Showing that $\frac{a + c}{b + d} \leq \max(\frac{a}{b},\frac{c}{d})$ I need help in proving (or disproving) the following assumption:

$$\frac{a + c}{b + d} \leq \max(\frac{a}{b},\frac{c}{d})$$

where $a,b,c,d \geq 0$ are positive integers. Both fractions $\frac{a}{b}$ and $\frac{c}{d}$ are between 0 and 1 and therefore the conditions $a \leq b$ and $c \leq d$ hold.
Any help or ideas are appreciated, thank you!
 A: Assume without loss of generality that $\dfrac{c}{d}\geq\dfrac{a}{b}$, i.e., $bc\ge ad$.
You want to show that $\dfrac{c}{d}-\dfrac{a+c}{b+d}\ge 0$. This means $\dfrac{bc-ad}{d(b+d)}\ge 0$. But this is true from the above.
The inequality can be generalized to more variables:
$$\frac{a_1+\dots+a_n}{b_1+\dots+b_n}\le \max\left(\frac{a_1}{b_1},\dots,\frac{a_n}{b_n}\right).$$
A: Holders inequality (the $p=\infty$,$q=1$ case) has:
$$a+c =\frac{a}{b} b + \frac{c}{d} d \leq \max\{\frac{a}{b},\frac{c}{d} \}b+\max\{\frac{a}{b},\frac{c}{d} \}d = \max\{\frac{a}{b},\frac{c}{d} \}\left(b+d\right)$$
Nonnegativity is needed in the step with the inequality and in dividing by $(d+b)$.
A: We wish to show that the inequality holds, with equality only when $a/b=c/d$
Consider the function $f(t)=(c/d)t+(a/b)(1-t)$.  This is a linear function equal to $a/b$ when $t=0$ and $c/d$ when $t=1$.  Moreover, $f(t)$ is between these two quantities if and only if $0\leq t\leq 1$.  Therefore, if we can show that $f(t)=(a+c)/(b+d)$ for some $t$ between $0$ and $1$, we will be done.
Let us solve:
$$(a+c)/(b+d)=(c/d)t+(a/b)(1-t)$$
Multiplying by $b*d$, expanding out, and rearranging terms, we get
$$d(bc-ad)/(b+d)=(bc-ad)t$$
Assuming that $bc=ad$ is nonzero (which happens when $a/b$ and $c/d$ are not equal), we can divide out to get $t=d/(b+d)$.  Assuming that $d>0$ and $b>0$, or that $b<0$ and $d<0$ we have $t$ is between $0$ and $1$, and we are done.  However, if the signs of $b$ and $d$ are different, then $t$ will be outside the range, and the inequality will not hold.
A: Without loss of generality, assume a/b ≤ c/d. Then ad ≤ bc and ad +cd ≤ bc +cd so that (a+c)d ≤ (b+d)c. It follows that (a+c)/(b+d) ≤ c/d.
There is no need to assume that the fractions are between 0 and 1. A similar argument shows that (a+c)/(b+d) lies between the min and the max of a/b, c/d. 
Google "Farey addition" and you will see that there is much more to adding fractions the way you always wanted to than you would guess.
