I have encountered a simple problem in my research, but I cannot solve it. we need not to have much knowledge to think this problem.
Let $1<\beta_1\,,\beta_2<2$ and $\beta_1\beta_2\leq2$ , set $$a=\frac{b(\beta_1\beta_2+\beta_2^2+\beta_1^2)}{(\beta_1+\beta_2)\beta_1\beta_2}$$ and $$c=\frac{b\beta_1\beta_2}{\beta_1+\beta_2}$$, then choose a,b,c such that $$|a-\beta_1-\beta_2|+|b-a(\beta_1+\beta_2)+\beta_1\beta_2|+|c\beta_1\beta_2|\leq2$$
First, I would rewrite it a little bit. Set $\alpha=\beta_1+\beta_2$ and $\gamma=\beta_1\beta_2$. Then you have \begin{align} a&=b\frac{\alpha^2-\gamma}{\alpha\gamma}\\ c&=b\frac{\gamma}{\alpha} \end{align} and your inequality becomes $$ |a-\alpha|+|b-a\alpha+\gamma|+|c\gamma|\leq 2 $$ which becomes \begin{align} \left|b\frac{\alpha^2-\gamma}{\alpha\gamma}-\alpha\right|+ \left|b-b\frac{\alpha^2-\gamma}{\gamma}+\gamma\right|+ \left|b\frac{\gamma^2}{\alpha}\right|\leq 2\\ \left|\frac{\alpha^2-\gamma}{\alpha\gamma}\right|\left|b-\frac{\alpha^2\gamma}{\alpha^2-\gamma}\right|+ \left|\frac{2\gamma-\alpha^2}{\gamma}\right|\left|b-\frac{\gamma^2}{2\gamma-\alpha^2}\right|+ \left|\frac{\gamma^2}{\alpha}\right||b-0|\leq 2\\ \end{align} On the left you have a sum of functions of the form $m|b-n|$, which is a modified absolute value function with a zero at $n$ and slope $\pm m$. It's pretty awkward to give a general solution for the interval of possible $b$'s depending on $\alpha$ and $\gamma$, but if you have concrete values given, it should be fairly easy to determine, since you have a convex piecewise linear function on the left.
EDIT: As we've already seen, the function on the left hand side takes on the form: $$f(b) = \sum_i m_i |b_i-n_i|$$ It's obvious that $f'(b)$ is non-decreasing, where it's defined (please plot an example to get a feel for it), so the function is convex. Further it's piecewise linear with kinks at the $n_i$. Hence, the minimum of this function is either at exactly at one of the $n_i$ or on a whole interval $[n_i, n_{i+1}]$. Thus it's enough to check the value of $f$ at each of the $n_i$ and take the minimum thereof. If $f(n_i)$ is smaller or equal to 2 for one $n_i$, than it's a solution, otherwise there is no solution at all. In your case you have to check three values: \begin{align} f\left(\frac{\alpha^2\gamma}{\alpha^2-\gamma}\right) &= \left|\frac{2\gamma-\alpha^2}{\gamma}\right|\left|\frac{\alpha^2\gamma}{\alpha^2-\gamma}-\frac{\gamma^2}{2\gamma-\alpha^2}\right|+ \left|\frac{\gamma^2}{\alpha}\right|\left|\frac{\alpha^2\gamma}{\alpha^2-\gamma}-0\right|\\ &= \left|\frac{\alpha^2(2\gamma-\alpha^2)}{\alpha^2-\gamma}-\gamma\right|+ \left|\frac{\alpha\gamma^3}{\alpha^2-\gamma}\right|\\ f\left(\frac{\gamma^2}{2\gamma-\alpha^2}\right) &= \left|\frac{\alpha^2-\gamma}{\alpha\gamma}\right|\left|\frac{\gamma^2}{2\gamma-\alpha^2}-\frac{\alpha^2\gamma}{\alpha^2-\gamma}\right|+ \left|\frac{\gamma^2}{\alpha}\right|\left|\frac{\gamma^2}{2\gamma-\alpha^2}-0\right|\\ &= \left|\frac{(\alpha^2-\gamma)\gamma}{\alpha(2\gamma-\alpha^2)}-\alpha\right|+ \left|\frac{\gamma^4}{\alpha(2\gamma-\alpha^2)}\right|\\ f\left(0\right) &= \left|\frac{\alpha^2-\gamma}{\alpha\gamma}\right|\left|0-\frac{\alpha^2\gamma}{\alpha^2-\gamma}\right|+ \left|\frac{2\gamma-\alpha^2}{\gamma}\right|\left|0-\frac{\gamma^2}{2\gamma-\alpha^2}\right|\\ &= \left|\alpha\right|+ \left|\gamma\right|\\ \end{align} If you have some $\beta_1$ and $\beta_2$ given, it should be fairly easy to find the minimum now. If you want to have a general idea, where the inequality is satisfied I would suggest plotting $\min(f(n_1), \ldots, f(n_3))$ over the allowed range of $\beta_1$ and $\beta_2$. Solving the inequality directly for those $\beta_1$ and $\beta_2$ that have a solution for $b$ seems way too complicated to me.
