Find the greatest possible real value of $S$ and smallest possible value of $T$ such that for every triangle with sides $a,b,c$ $(a\le b\le c)$ to be true the inequalities:
$$S\le\frac{(a+b+c)^2}{bc}\le T.$$
1. We start with the given inequality:
   \[
   S \le \frac{(a+b+c)^2}{bc} \le T
   \]
   where \(a \le b \le c\) are the sides of a triangle.

2. We rewrite the expression inside the inequality:
   \[
   \frac{(a+b+c)^2}{bc}
   \]
   This can be expanded as:
   \[
   \frac{a^2 + b^2 + c^2 + 2ab + 2bc + 2ca}{bc}
   \]

3. We use the fact that in any triangle, the sides satisfy the triangle inequality. Specifically, \(b + c \ge a\), \(a + c \ge b\), and \(a + b \ge c\).

4. To find the maximum value of \(\frac{(a+b+c)^2}{bc}\), we consider the case when \(a\) is very small compared to \(b\) and \(c\). In this case, \(a \approx 0\), and the expression simplifies to:
   \[
   \frac{(b+c)^2}{bc}
   \]
   Using the Arithmetic Mean-Geometric Mean (AM-GM) inequality, we have:
   \[
   b + c \ge 2\sqrt{bc}
   \]
   Squaring both sides, we get:
   \[
   (b+c)^2 \ge 4bc
   \]
   Therefore:
   \[
   \frac{(b+c)^2}{bc} \ge \frac{4bc}{bc} = 4
   \]
   This shows that the minimum value of \(\frac{(a+b+c)^2}{bc}\) is 4.

5. To find the minimum value of \(\frac{(a+b+c)^2}{bc}\), we consider the case when \(a\), \(b\), and \(c\) are all equal. Let \(a = b = c\). Then:
   \[
   \frac{(a+a+a)^2}{a \cdot a} = \frac{(3a)^2}{a^2} = \frac{9a^2}{a^2} = 9
   \]
   This shows that the maximum value of \(\frac{(a+b+c)^2}{bc}\) is 9.

6. Therefore, the greatest possible real value of \(S\) is 4, and the smallest possible value of \(T\) is 9.

The final answer is \( \boxed{ S = 4 } \) and \(T = 9\).