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
 A: 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
$$
\frac{1}{\frac{a}{x}+\frac{b}{y}}\le ax+by
$$
A: 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.
A: This is just Cauchy-Schwarz:
$$1=(a+b)^2=\left(\sqrt{ax} \sqrt{\frac ax} + \sqrt{by} \sqrt{\frac by}\right)^2 \le (ax+by)\left(\frac ax + \frac by\right)$$
A: 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.
