If $\|x\|\leq \|y+z\|$ is $\|\alpha x\| \leq \|y+\alpha z\|$ for $\alpha\in[0,1]$?

Question

If $\|x\|\leq \|y+z\|$ is $\|\alpha x\| \leq \|y+\alpha z\|$ for $\alpha\in[0,1]$? For reference, this is not a homework problem, but it did come up in something I'm looking at. If $\langle y,z\rangle\geq 0$, the proof seems pretty straightforward. Simply \begin{align*} \|\alpha x\|^2 =& \alpha^2\|x\|^2\\ \leq&\alpha^2 \|y+z\|^2\\ =&\alpha^2 \|y\|^2 + 2 \alpha^2\langle y,z\rangle+\alpha^2\|z\|^2\\ \leq&\|y\|^2 + 2 \alpha^2\langle y,z\rangle+\alpha^2\|z\|^2\\ \leq&\|y\|^2 + 2 \alpha\langle y,z\rangle+\alpha^2\|z\|^2\\ =&\|y+\alpha z\|^2 \end{align*} However, when $\langle y,z\rangle < 0$, I'm not so sure.

Answer 1

If $\alpha=0,1$, the solution is obvious. Now lets consider $\alpha\in(0,1)$. However, for \begin{align} x=\begin{pmatrix} 1\\0 \end{pmatrix}, y= \begin{pmatrix} -1\\-1 \end{pmatrix}, z= \begin{pmatrix} 2\\2 \end{pmatrix}, \alpha=\frac 1 2 \end{align} we have $$\|x\|=1\leq\sqrt 2=\|y+z\|$$ and $$\|y+\alpha z\|=0<\frac 1 2=\|\alpha x\|.$$ Thus, you're going to need some stronger conditions for your inequality to hold in general. Perhaps some relation on the $x,y,z$?