i need to show that$$\left|\frac{1}{a^2}-\frac{1}{b^2}\right|\leq\frac{2|a-b|}{k^3}$$where $a,b,k\in\mathbb{R}$, $|a|k0$ and $|b|k0$. ive tried a few things, but im stuck. help!

Question

i need to show that$$\left|\frac{1}{a^2}-\frac{1}{b^2}\right|\leq\frac{2|a-b|}{k^3}$$where $a,b,k\in\mathbb{R}$, $|a|>k>0$ and $|b|>k>0$. ive tried a few things, but im stuck. help!

jhonyy9 · Answer

- so what do you think from this ?

the first term will be k2*I(b2-a2)/a2b2)I <= 2Ia-bI
-k2*I(a-b)(a+b)I<=2Ia2b2*(a-b)I

anonymous · Answer

i tried$$\left|\frac{1}{a^2}-\frac{1}{b^2}\right|=\left|\frac{b^2-a^2}{(ab)^2}\right|=\frac{|a^2-b^2|}{(ab)^2}=\frac{|(a-b)(a+b)|}{(ab)^2}$$