The solving set of $2a+b=3c$

Question

Is there a possibility where $a>b>c$ in the equation of $2a+b=3c$ and where $b>a>c$ in the equation of $2a+b=3c$?

Answer 1

Another way of seeing this, more generally, is:

If $a\neq b$ are real numbers and $0<t<1$ then $ta+(1-t)b$ is strictly between $a$ and $b.$

Proof:

$a\leq \max(a,b). b\leq \max(a,b)$, with one of the two being strict inequality. So, since $t>0$ and $1-t>0,$ you have that $ta+(1-t)b<t\max(a,b)+(1-t)\max(a,b)=\max(a,b).$

Similarly, $ta+(1-t)b>\min(a,b).$

In your question: $$3c=2a+b\implies c=ta+(1-t)b, t=\frac{2}{3}$$ and $$3c=a+2b\implies c=ta+(1-t)b, t=\frac{1}{3}$$

So in both cases, if $a\neq b$ then $c$ must be strictly between $a,b.$

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Another fun approach. Let $f(x)=(x-a)(x-b).$ Then if $f(x)<0$, then $x$ is strictly between $a$ and $b$, if $f(x)=0$ then $x=a$ or $x=b$. Otherwise, $f(x)>0.$

Now, if $3c=2a+b$ then $$f(c)=\left(\frac{2a+b}{3}-a\right)\left(\frac{2a+b}{3}-b\right)=\frac{b-a}{3}\cdot\frac{2(a-b)}{3}=-\frac{2(a-b)^2}{9}<0$$

if $a\neq b,$ and $f(c)=0$ if $a=b.$

More generally, our above lemma can be proved this same way since, if $c=ta+(1-t)b$ with $0<t<1,$ then:

$$f(c)=(1-t)\frac{b-a}{2}\cdot t\frac{a-b}{2}=-t(1-t)(b-a)^2$$ and $t(1-t)>0.$

Answer 2

Note that

$$a>b>c \implies 2a+b>3b>3c$$

and

$$b>a>c \implies 2a+b>3a>3c$$