Does this norm inequality always hold? Let $(X,\|\cdot\|)$ be a normed vector space.
Let $x,y,z\in X$ with $\|x-y\|,\|x-z\|\leq\frac12\|x\|$.
Define $y':=\left(1+\frac{\|x-y\|}{\|y\|}\right)y$ and define $z'$ similarly.
Does $\|y'+z'\|\geq2\|x\|$ always hold?
More generally, does $\|\lambda y'+\mu z'\|\geq(\lambda+\mu)\|x\|$ always hold?
If you make drawings of the situation in $\mathbb{R}^2$ it seems like this should always hold, but I can not find any proof.
 A: The inequality does always hold in a real inner product space. More generally, instead of $\|x-y\|,\|x-z\|\leq\frac12\|x\|$, we can require $\langle x,y\rangle,\langle x,z\rangle\geq\frac12\|x\|^2$.
Lemma:
For $a,b,c\in\mathbb{R}$ with $a\leq b$ and $b>0$ we have $$\frac{\sqrt{a^2+c^2}}{\sqrt{b^2+c^2}}\geq\frac{a}b.$$
Proof of lemma:
If $a<0$, then the left hand side is positive while the right hand side is negative.
If $a\geq0$, then $a^2\leq b^2$ and $\frac{a^2}{b^2}\leq1$, so $$\frac{a^2+c^2}{b^2+c^2}=\frac{b^2}{b^2+c^2}\cdot\frac{a^2}{b^2}+\frac{c^2}{b^2+c^2}\cdot1\geq\frac{b^2}{b^2+c^2}\cdot\frac{a^2}{b^2}+\frac{c^2}{b^2+c^2}\cdot\frac{a^2}{b^2}=\frac{a^2}{b^2}.$$
Proof of statement:
Let us assume without loss of generality that $\|x\|=1$.
Let $y=\alpha x+p$, $y'=\alpha' x+p'$, $z=\beta x+q$ and $z'=\beta' x+q'$ with $p,p',q,q'\perp x$.
Then $\|y\|=\sqrt{\alpha^2+\|p\|^2}$ and $\|x-y\|=\sqrt{(1-\alpha)^2+\|p\|^2}$.
Note that $\alpha=\langle x,y\rangle\geq\frac12$, so $\alpha>0$ and $\alpha\geq1-\alpha$.
From the lemma, it follows that $\alpha'=\left(1+\frac{\|x-y\|}{\|y\|}\right)\alpha\geq1$.
Similarly, $\beta'\geq1$.
Hence $\|y'+z'\|=\sqrt{(\alpha'+\beta')^2+\|p'+q'\|^2}\geq\alpha'+\beta'\geq2$.
More generally, $\|\lambda y'+\mu z'\|=\sqrt{(\lambda\alpha'+\mu\beta')^2+\|\lambda p'+\mu q'\|^2}\geq\lambda\alpha'+\mu\beta'\geq\lambda+\mu$.
A: Third time's the charm. I think it's false even when $z=x$. Let $(X,||\cdot||) = (C([0,1]),||\cdot||_\infty)$. 
We have $||x||_\infty = 1, ||y||_\infty = \frac{3}{4}, ||x-y||_\infty = \frac{1}{2}$, and $||(1+\frac{1/2}{3/4})y+x||_\infty = ||\frac{5}{3}y+x||_\infty = \frac{11}{6} < 2 = 2||x||_\infty$.

