# Inequality involving Harmonic mean

by all.over   Last Updated October 10, 2019 03:20 AM - source

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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Updated January 19, 2018 23:20 PM