# 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.

• Have you tried checking it for some values, e.g. $a = 1, b = 2, c = 3, d = 4$? Commented Sep 26, 2013 at 18:41
• Or a = b = c = d = 1... Commented Sep 26, 2013 at 18:42
• Daniel: it was typing mistake Commented Sep 26, 2013 at 18:45
• You will find your answer here: en.wikipedia.org/wiki/Mediant_(mathematics) Commented Sep 26, 2013 at 18:45
• @Thomas, the version of the inequality in the title and first line moves the letter $b$ from the numerator to the denominator. That is not how the mediant works. It also moves $c$ from the denominator to the numerator. Commented Sep 26, 2013 at 19:48

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:

$$\frac{a_1+\dots+a_n}{b_1+\dots+b_n}\leq \max_i\left(\frac{a_i}{b_i}\right)$$

• @Phani, notice that what you typed in your title is completely unrelated to what Thomas answered. Clearly, since you accepted it, this is what you meant. Learn to proofread what you post, including the title!!! Commented Sep 26, 2013 at 19:12
• I edited in a mediant version of the question in magenta. Commented Sep 26, 2013 at 20:16
• @WillJagy yes, I didn't read carefully, assume it was the simple and obvious question. Commented Sep 26, 2013 at 21:57
• I read the math symbols very carefully for up to two lines, if there is more than that I go look for other questions. I mostly ignore the words. Commented Sep 26, 2013 at 22:02

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.

Let $$a=\lambda c$$ and $$b=\mu d$$, where $$\lambda, \mu>0$$. $$\frac {a+b}{c+d}=\frac {\lambda c+\mu d}{c+d}=\frac {\lambda (c+d)+\mu (c+d)-(\lambda d+\mu c)}{c+d}=\lambda + \mu -\frac {\lambda d+\mu c}{c+d}<\lambda + \mu = \frac ac+\frac bd$$