Inequality involving Harmonic mean

Question

Let $a, b \ge 0 $ and $x,y > 1$ Show $\frac{1}{1/x+1/y}(a+b) \le \max(ax,by)$

If $a = b = 0$ then this is clear so assume not both are zero.

This seems to be related to the harmonic mean but I am not quite getting it.

$\frac{1}{1/x+1/y}(a+b) = \frac{xy}{x+y}(a+b) = \frac{1}{2}H(x,y)(a+b)$

Where $H(x,y) = \frac{2xy}{x+y}$ : the harmonic mean of $x,y$

And, $\max(ax,by) = \frac{1}{2}[ax+by+\mid ax - by \mid ] = A(ax,by) + \frac{1}{2}\mid ax - by \mid$

Where $A(ax,by) = \frac{ax+by}{2}$ : the arithmetic mean of $ax,by$

This is where I get stuck.

EDIT: I am closer but not quite:

$\displaystyle \frac{a+b}{x+y} = \frac{a}{x+y}+\frac{b}{x+y} \le \frac{a}{y} + \frac{b}{x} \implies xy\frac{a+b}{x+y} \le xy(\frac{a}{y} + \frac{b}{x} ) = ax + by \le 2\max(ax,by) $

Answer 1

In fact, the conditions for $x,y,a,b$ can be loosened:

For all $x,y > 0$ and $a,b \in \mathbb{R}$, we have $$\frac{1}{1/x+1/y}(a+b)=\dfrac{y\cdot ax+x\cdot by}{x+y} \le \dfrac{y\cdot \max(ax,by)+x\cdot \max(ax,by)}{x+y} =\max(ax,by).$$

Answer 2

Cauchy Schwarz says for $a,b\in\mathbb{R}$ and $x,y\gt0$, $$ (a+b)^2\le\left(a^2x+b^2y\right)\left(\frac1x+\frac1y\right) $$ which gives, for $a+b\ge0$, $$ \begin{align} \frac1{\frac1x+\frac1y}(a+b) &\le\frac{a(ax)+b(by)}{a+b}\\ &\le\max(ax,by) \end{align} $$