Explain me, please

Question

Explain me, please

miracrown · Answer

Explain you?

kainui · Answer

3 points away from being a winner.

loser66 · Answer

$$\mid \dfrac{x_0+x}{x^2x_0^2} \mid$$
where $$\dfrac{3x_0}{2}<\mid x_0+x\mid<\dfrac{5x_0}{2}$$

loser66 · Answer

and $$\dfrac{x_0^4}{4}<x^2x_0^2<\dfrac{9x_0^2}{4}$$

loser66 · Answer

how to combine them to get $\mid \dfrac{x+x_0}{x^2x_0^2}\mid< \dfrac{10}{x_0^3}$

rsadhvika · Answer

$$ \left| \dfrac{x_0+x}{x^2x_0^2} \right| =   \dfrac{\left|x_0+x\right|}{\left|x^2x_0^2\right|}   $$

rsadhvika · Answer

mess with that fraction
you know min/max values for both numerator and denominator

miracrown · Answer

First of all, we should assume x_0 is positive. Otherwise, the first inequality is trivial on the left side...actually it's impossible

loser66 · Answer

@Kainui  I am naturally a big LOSER, 3 more points away is not a big deal to me, hehehe

miracrown · Answer

absolute value is always nonnegative, so that x_0 must be nonnegative also 
positive because we're putting it in the denominator

loser66 · Answer

@Miracrown  got you, then?

miracrown · Answer

So let's see how we can go about proving this ...

ikram002p · Answer

miracrown · Answer

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