Existence of a "scaling factor" in a inequality

Question

We're given the real numbers $a,b,c$ that satisfy the condition $a\le b \le c$.

Consider the case where $b\le \frac{a+b+c}{3}$, then we have $\frac{a+b+c}{3} \le \frac{a+c}{2} \le c$ and similarly $\frac{a+b+c}{3} \le \frac{b+c}{2} \le c$, Then there exist $\lambda ,\mu \in [0,1]$ such that: $$\frac{c+a}{2}=\lambda c+(1-\lambda)\left(\frac{a+b+c}{3}\right)$$ and $$\frac{b+c}{2}=\mu c+(1-\mu)\left(\frac{a+b+c}{3}\right)$$ I can "intuitively" see why this should be true, but why do such $\lambda$ and $\mu$ must exist?

Answer 1

With $x = \frac{a+b+c}{3}, y= \frac{a+c}{2}, z= c$ your question becomes:

If $x \le y \le z$ then there is a $\lambda \in [0, 1]$ such that $$ \tag{*} y = \lambda x + (1 - \lambda) z \, . $$

If $x = z$ then any $\lambda \in [0, 1]$ will do, otherwise $(*)$ is equivalent to $$ \lambda = \frac{z-y}{z-x} \, , $$ which satisfies $0 \le \lambda \le 1$.