# Prove that the ratio $la+mc+ne : lb+md+nf$ will be equal to each of the ratios $a:b, c:d, e:f$

Prove that the ratio $$la+mc+ne : lb+md+nf$$ will be equal to each of the ratios $$a:b, c:d, e:f$$, if these be all equal; and that it will be intermediate in value between the greatest and least of these ratios if they be not all equal.

If all the ratios are equal: $$\dfrac{a}{b}=\dfrac{c}{d}=\dfrac{e}{f}=k$$

Then $$a=bk, c=dk, e=fk$$

$$\Rightarrow\dfrac{la+mc+ne}{lb+md+nf}=\dfrac{lbk+mdk+nfk}{lb+md+nf}=\dfrac{k(lb+md+nf)}{lb+md+nf}=k$$

If all the ratios are not equal, prove that $$la+mc+ne : lb+md+nf$$ will be intermediate in value between the greatest and least of these ratios.

I don't know how to do this. I tried $$\dfrac{a}{b}=x,\ \ \dfrac{c}{d}=y,\ \ \dfrac{e}{f}=z$$, (Since the ratios are not equal)

Then $$\dfrac{lbx+mdy+nfz}{lb+md+nf}$$. But this looks wrong because I can't do anything with it. I also have no idea how to determine which is greatest and least among $$\dfrac{a}{b},\ \ \dfrac{c}{d},\ \ \dfrac{e}{f}$$

Thanks for the help.

• Are there sign hypothesis on $a,b,\dots,m,n?$ If the three numbers $lb,md,nf$ are $>0,$ assuming wlog $x\le y\le z,$ you easily get$$x\le\dfrac{lbx+mdy+nfz}{lb+md+nf}\le z.$$ Commented Jun 7, 2023 at 10:39
• Thanks for the help. My question is how can you write $lbx+mdx+nfx$ or $lbz+mdz+nfz$? Because the original statement is $la+mc+ne$, and knowing the ratios are not equal to each other then shouldn't the only acceptable statement be $la+mc+ne=lbx+mdy+nfz$? Because to me $lbx+mdx+nfx$ or $lbz+mdz+nfz$ means the ratios are all equal to $x$ or $z$. Sorry if this question sounds amateurish but this is what I'm thinking. Thanks. Commented Jun 7, 2023 at 13:29
• In your $\dfrac{lbx+mdy+nfz}{lb+md+nf}$, I took an upper (resp. lower) bound (not $=$) of the numerator by replacing $x,y$ with $z$ (resp. $y,z$ with $x$). Commented Jun 7, 2023 at 13:34

Your last expression isn't wrong, it's on the contrary.

$$\dfrac{lb\min(x,y,z)+md\min(x,y,z)+nf\min(x,y,z)}{lb+md+nf} = \min(x,y,z) \le \dfrac{lbx+mdy+nfz}{lb+md+nf} \le \dfrac{lb\max(x,y,z)+md\max(x,y,z)+nf\max(x,y,z)}{lb+md+nf} = \max(x,y,z)$$

As @Anne Bauval pointed out, I also assumed all variables to be positive.

Assume that the three numbers $$lb,md,nf$$ are $$>0$$ (else, you will easily find a counterexample).

You do not need to "determine which is greatest and least among $$\dfrac{a}{b},\ \ \dfrac{c}{d},\ \ \dfrac{e}{f}$$": as previously commented, you can assume without lost of generality that$$x\le y\le z$$(by some permutation on the three $$4$$-tuples $$(a,b,x,l),(c,d,y,m),(e,f,z,n)$$). This has the advantage of simplifying the notations, avoiding the use of expressions involving $$\min$$ or $$\max.$$

Then, you easily get$$x\le\dfrac{lbx+mdy+nfz}{lb+md+nf}\le z.$$ For this, just notice that the numerator is $$\ge lbx+mdx+nfx$$ and $$\le lbz+mdz+nfz.$$

• In my answer I avoided the "determine which is greatest and least", while you chose the "assuming wlog" route, however I guess the OP doesn't know why we can assume that, otherwise they wouldn't want to determine, so that could be added to the answer. Commented Jun 7, 2023 at 11:00