# Given positive real numbers a and b, prove that $\frac{2}{\frac{a}{x}+\frac{b}{y}} \leq ax + by, x>0, y>0$

I am working through Problem Solving through Problems by Larsons. I am stuck on question 1.8.5b) which is as follows:

" Given positive real numbers $$a$$ and $$b$$ such that $$a + b = 1$$, prove that

$$\frac{2}{\frac{a}{x}+\frac{b}{y}} \leq ax + by, \ x>0, y>0$$

"

It appears to me that this question has a typo. Given any $$a$$ and $$b$$, choose $$x = 1$$, $$y = 1$$. Then:

$$\frac{2}{\frac{a}{1}+\frac{b}{1}} \leq a + b \Leftrightarrow \frac{2}{a+b} \leq a + b \Leftrightarrow \frac{2}{1} \leq 1 \Rightarrow \Leftarrow Contradiction$$

My long-shot hypothesis is that the "$$2$$" is supposed to be a "$$1$$". I think that I actually proved this version of the problem (it was how I first realized that something seemed weird in the problem -- won't get into that proof here). However, I am hugely unsure about all of this. It seems very bizarre to me that Problem Solving would have this sort of error, and I suspect that I am just getting something very simple wrong somewhere. Anyone agree with my reasoning or know where I went wrong? Thank you!

Edit: Thought I would drop a link to the original text where the problem appears as above: https://math.la.asu.edu/~ifulman/spring13/mat194/problem-solving.pdf

• hm Interesting... Commented Dec 21, 2020 at 1:09
• There isn't an answers section in that book? Commented Dec 21, 2020 at 1:14
• @Gerry Myerson Haha unfortunately not. The book is actually fairly infamous for that.. Commented Dec 21, 2020 at 1:19
• OK. Anyway, I'm glad you got an answer from FFjet. Commented Dec 21, 2020 at 1:25

You are right, the problem is wrong. Actually we have $$\left( \frac{a}{x}+\frac{b}{y} \right) \left( ax+by \right) =a^2+b^2+ab\left( \frac{y}{x}+\frac{x}{y} \right) \ge a^2+b^2+2ab=\left( a+b \right) ^2=1$$ which indicates $$\frac{1}{\frac{a}{x}+\frac{b}{y}}\le ax+by$$

• Yes, that was precisely my proof! Thanks for the response. Commented Dec 21, 2020 at 1:18

Note $$\frac{1}{\frac{a}{x}+\frac{b}{y}} \leq ax + by\tag1$$ is equivalent to $$(ax+by)(\frac{a}{x}+\frac{b}{y})\ge1$$ or $$a^2+b^2+ab(\frac{x}{y}+\frac{y}{x})\ge1. \tag2$$ Since $$\frac{x}{y}+\frac{y}{x}\ge 2$$ one has $$a^2+b^2+ab(\frac{x}{y}+\frac{y}{x})\ge a^2+b^2+2ab=(a+b)^2=1$$ which says (2) is true. So (1) is true.

This is just Cauchy-Schwarz:

$$1=(a+b)^2=\left(\sqrt{ax} \sqrt{\frac ax} + \sqrt{by} \sqrt{\frac by}\right)^2 \le (ax+by)\left(\frac ax + \frac by\right)$$

Actually, you can get an identity rather than just an inequality.

I'll start like FFjet.

$$\begin{array}\\ \left( \frac{a}{x}+\frac{b}{y} \right) \left( ax+by \right) &=a^2+b^2+ab\left( \frac{y}{x}+\frac{x}{y} \right)\\ &=a^2+b^2+2ab +ab\left( \frac{y}{x}-2+\frac{x}{y} \right)\\ &=(a+b)^2+ ab\left( \sqrt{\frac{y}{x}}-\sqrt{\frac{x}{y}} \right)^2\\ &=(a+b)^2+ \frac{ab}{xy}\left(y-x\right)^2\\ &\ge (a+b)^2 \qquad\text{with equality } \iff x=y\\ \end{array}$$

So it doesn't depend on what $$a+b$$ is.