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) $

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