Does $3 (a+2c) =4 (b+3d)$, with $d>0$, imply $a+c>b+d$?

Question

If $3 (a+2c) =4 (b+3d)$ and $d>0$, then, is the inequality $a+c>b+d$ true? If so, prove.

Answer 1

If this true then since $d=\frac{3a+6c-4b}{12}$, we need $$a+c>b+\frac{3a+6c-4b}{12}$$ or $$9a+6c-8b>0$$ is true for all reals $a$, $b$ and $c$ such that $3a+6c-4b>0.$

Now, try to find a counterexample.

Take $b=0$ and $c=1$. What is a value of $a$ for counterexample?