Let $\displaystyle |x|=|y|=r$ with $\displaystyle x \neq y$. If $\displaystyle 0<t<1$ prove that $\displaystyle |(1-t)x+ty|<r$.

I've proven that $\displaystyle |(1-t)x+ty| \leq r$. Now how do I show that this inquality is strict?

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- Mar 19th 2011, 07:49 AMjefferson_lcopen ball
Let $\displaystyle |x|=|y|=r$ with $\displaystyle x \neq y$. If $\displaystyle 0<t<1$ prove that $\displaystyle |(1-t)x+ty|<r$.

I've proven that $\displaystyle |(1-t)x+ty| \leq r$. Now how do I show that this inquality is strict? - Mar 19th 2011, 09:01 AMPlato
- Mar 19th 2011, 09:11 AMjefferson_lc
- Mar 19th 2011, 11:34 AMOpalg
You need to give us a bit more context here. What are x and y supposed to be – real numbers, complex numbers, vectors in a euclidean space, or what?

In fact, in every case the method is the same: form the square $\displaystyle |(1-t)x+ty|^2$. Expand that, either as a product or in terms of inner products, depending on what sort of space you are working in, and see when that can be equal to $\displaystyle r^2$. You should end up with an equation like $\displaystyle \langle x,y\rangle = r^2$ (again, depending on the space). You then need to show why that can only happen when $\displaystyle x = y$.