Re: x < 1/y, and x and y are positive
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03 Apr 2019, 02:29
First, let me break up each quantity:
Quantity A:
\(\frac{2}{x}+\frac{x}{x}-\frac{x^{2}}{x} = \frac{2}{x}+1-x\)
Quantity B:
\(\frac{2y^{2}}{y}+\frac{y}{y}-\frac{1}{y} = 2y+1- \frac{1}{y}\)
Now subtract 1 from each quantity:
Quantity A: \(\frac{2}{x}-x\)
Quantity B: \(2y - \frac{1}{y}\)
Now we need some intuitive mathematical thinking. What if, instead of \(x < \frac{1}{y}\), we had \(x = \frac{1}{y}\)? In that case, both quantities would be equal. Quantity A would become \(\frac{2}{\frac{1}{y}} - \frac{1}{y} = 2y - \frac{1}{y}\), same as quantity B.
In reality, though, x is a bit less than \(\frac{1}{y}\). What happens when x gets smaller? Well, when x decreases, \(\frac{2}{x}\) increases, while the number we subtract, x, decreases. So when x gets smaller, quantity A gets larger.
Putting it all together, if x was equal to \(\frac{1}{y}\), then both quantities would be equal. But x is actually a bit less than that, and as x decreases, quantity A increases while quantity B stays the same. Hence, quantity A is larger, no matter what.