$x\perp y$ iff $\|x+\lambda y\|\geq \|x\|$ for all scalars $\lambda$

Question

Show that in an inner product space $X$

a) $x\perp y$ iff $\|x+\lambda y\|=\|x-\lambda y\|$ for all scalars $\lambda$

b) $x\perp y$ iff $\|x+\lambda y\|\geq \|x\|$ for all scalars $\lambda$

My Work:

I did part a). Did only the forward direction of part b). Can anybody please give me a hint for the other direction? Tried it using $\lambda=1,-1,i,-i$, but no success.

Answer 1

$$||x+\lambda y||\geq ||x||\iff ||x+\lambda y||^2\geq ||x||^2\iff P(\lambda)= 2Re[\overline{\lambda}\langle x,y\rangle] +|\lambda|^2||y||^2\ge0$$ If $y=0$ the answer is trivial. Assume therefore $y\neq 0$. Then, if we select $$\lambda_0=-\frac{\langle x,y\rangle}{\|y\|^2}$$ we have $$P(\lambda_0)=-\frac{|\langle x,y\rangle|^2}{\|y\|^2}$$ which is nonnegative only if $\langle x,y\rangle =0$.