Prove $||\lambda x_1 + (1-\lambda) x_2 - y|| \leq ||x_1 - y||$

Question

Assume we have have $3$ points $x_1, x_2$ and $y$ and $||x_1-y||=||x_2-y||$. How do we prove that the distance between $y$ and the convex combination of $x_1$ and $x_2$ is smaller than that between $x_1$ and $y$?

Answer 1

Here I provide a solution only for an inner product space $H$. Let $a=x_1-y$ and $b=x_2-y$. Then $\|a\|=\|b\|$. WLOG, set $\|a\|=\|b\|=1$ and hence $(a,b)\le 1$. If $(a,b)=1$, then $a=b$. If $(a,b)<1$, then for $\lambda\in [0,1]$, \begin{eqnarray} f(\lambda)&=&\|\lambda x_1+(1-\lambda)x_2-y\|^2\\ &=&\|\lambda a+(1-\lambda)b\|^2\\ &=&\lambda^2+(1-\lambda)^2+2\lambda(1-\lambda)(a,b)\\ &=&2\lambda^2(1-(a,b))-2\lambda(1-(a,b))+1\\ &=&2(1-(a,b))(\lambda^2-\lambda)+1 \end{eqnarray} reaches the minimum $\frac{1}{4}\|a-b\|^2$ at $\lambda=\frac{1}{2}$. So $f(\lambda)$ reaches the maximum at either $\lambda=0$ or $\lambda=1$. Note $f(0)=f(1)=\|a\|=\|b\|$ and hence $f(\lambda)\le |a|^2$ or $$ \|\lambda x_1+(1-\lambda)x_2-y\|\le \|x_1-y\|. $$

Answer 2

Apply the triangle inequality to $ ||\lambda(x_1-y)+(1-\lambda)(x_2-y)||$.

Answer 3

Assuming $0\leq \lambda \leq 1$, $$||\lambda x_1 + (1-\lambda)x_2 - y|| = ||\lambda x_1 + (1-\lambda)x_2 - (\lambda + (1-\lambda))y||$$$$\leq ||\lambda x_1 - \lambda y|| + ||(1-\lambda)x_2 - (1-\lambda)y||$$$$ = \lambda||x_1 - y|| + (1-\lambda)||x_2 - y||$$$$ = \lambda||x_1 - y|| + (1-\lambda)||x_1 - y|| = ||x_1 - y||$$