Finding a contradiction in equations and inequalities. Question:-In the figure shown, Coefficient of friction between the blocks C and B is $0.4$. There is no friction between block C and ground.The system of blocks is released from rest in the shown situtation.Find the accelerations of masses.(Given $m_{B}=5kg , m_{C}=10kg , m_{A}=3kg$ )

I made two cases
$1)$ Block B and C move together.
$2)$ Block B and C will not move together.
When I solve equations for Case $1$, I find that block will move together.This implies that Case $1$ is true.
$$30-T=3a_{3}$$
$$T-f=5a_{5}$$
$$f=10a_{10}$$
$$f \leq 20$$
Note that if blocks B and C will move together then $a_{3}=a_{5}=a_{10}=a$
On solving we get $a=\frac {5}{3}m/s^2$ and $f=16.67N$
As case $1$ is true. So contradiction condition must arrive during solving equations for case $2$.
I got the following equations for case $2$.  Despite my all efforts I was unable to find how equations will contradict.
\begin{align}
30-T&=3a_{3}\\
T-f&=5a_{5}\\
f&=10a_{10}\\
f&\leq {20}\\
a_{3}&>a_{10}\\
\end{align}
Note that $a_{3}=a_{5}$
My attempt
Adding equation $1$ and $2$, we get
$$30-f=8a_{3}$$
So, $30-8a_{3}\leq 20$ and $10a_{10}\leq 20$
This gives $a_{3}\geq \frac54$ and $a_{10}\leq 2$
Also, $$T-f \geq \frac{25}{4}$$
$$30-T\geq \frac{15}{4}$$
This gives $T\leq \frac{105}{4}$
Adding eqation $1$ and $3$ we get
$$30+f-T=10a_{10}+3a_{3}$$
Using $f-T\leq \frac {-25}{4}$
$$12a_{3}+40a_{10}\leq {95}$$
Adding equation $2$ and $3$,we get
$$T=5a_{3}+10a_{10}$$
Using $T\leq \frac {105}{4}$
$$4a_{3}+8_{10}\leq 21$$
So We got following inequalities involving $a_{3}$ and $a_{10}$

$$a_{3}>0$$
$$a_{10}>0$$
$$a_{3}>a_{10}$$
$$a_{3}\geq \frac54$$
$$a_{10}\leq 2$$
$$12a_{3}+40a_{10}\leq {95}$$
$$4a_{3}+8a_{10}\leq 21$$

On plotting them on graph where x-axis represents $a_{3}$ and y-axis represents $a_{10}$ I got quadrilateral having coordinates $(1.25,0), (1.25,1.25), (1.75,1.75), (5.25,0)$.
So, $1.25\leq a_{3} \leq 5.25$ and $0\leq a_{10}\leq 1.75$
Can anybody help to find contradiction.
 A: I belive you have made a mistake. In either case we can calculate the frictional force directly, since we have friction at a maximum as we have motion:
$$F_\text{MAX}=\mu R=0.4\times5g=2g$$
I will calculate all accelerations below:
For B and A, let $a_1$ be their common acceleration. Resolve for A and then B:
$$\begin{align}
& 3g-T=3a_1\\
& T-2g=5a_1
\end{align}$$
Add the two equations together to eliminate $T$:
$$g=8a_1\implies a_1=\frac{g}{8}$$
This is actually equivalent to treating A and B as a single particle:
Using $F=ma$, it is clear that as the tensions cancel out we have
$$3g-2g=(3+5)a_1\implies g=8a_1\implies a_1=\frac{g}{8}$$
Now for box C. We have
$$2g=10a_2\implies a_2=\frac{g}{5}$$
where $a_2$ is the acceleration of C.
As an aside, we can prove that there is motion as follows:
First, note that the tension must be less than $3g$; otherwise A would move upwards, which is absurd. Now suppose that $T\le2g$. Then A will still accelerate downwards. But $F_\text{MAX}=2g$, so B will not move. This is a contradiction as B is connected to A. Hence $2g<T<3g$ and there is movement since $T>F_\text{MAX}$ as required.
