Proving this inequality

Question

I am having trouble with proving an inequality. Assume we have two positive real numbers $a$ and $b$ such that $a+b=1$ and numbers $x > 0$ and $y > 0$. Prove:

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

I'd appreciate any hints!

Answer 1

If you change 2 by 1 then the inequality holds

$1 \leq a^2+b^2 + ab(\frac{x}{y} + \frac{y}{x}) $

$1 \leq (a+b)^2 + ab(\frac{x}{y} + \frac{y}{x})-2ab$

$1 \leq 1^2 + ab(\frac{x}{y} + \frac{y}{x}-2)$

$0 \leq ab(\frac{x^2 + y^2 - 2xy}{xy}) $

$0 \leq ab(\frac{(x-y)(x-y)}{xy})$

$0 \leq ab(\frac{(x-y)^2}{xy})$

as $ab \geq 0$ , $(x-y)^2 \geq 0 $ and $xy > 0 $

Answer 2

If you multiply out, you get $2 \le a^2+b^2+ab(\frac{x}{y}+\frac{y}{x})$. Do you know of an inequality to get a lower bound for the right-hand expression? By the way, sometimes it is useful to homogenize the equation by using the constraint, in this case we would multiply $(a+b)^2$ to the left-hand expression.

Answer 3

The assertion is not true, consider the following counter example:$x=y=1 \implies 2\le 1$,perhaps "$2$" should be "$1$"

Answer 4

the function $f(x) = \dfrac{1}{x}$ is convex then : $$a f(x) + b f(y) \geq f(a x + b y)$$ We deduce that : $$\dfrac{a}{x} + \dfrac{b}{y} \geq \dfrac{1}{a x + b y}$$ hence : $$\dfrac{1}{\dfrac{a}{x} + \dfrac{b}{y}} \leq a x + b y$$