Let $a \leq b \leq c$ be the sides of a triangle. What values can

$$
\frac{(a+b+c)^{2}}{b c}
$$

take?

OKTV, 1995-96
First, we show that under our conditions,

$$
4\frac{(b+c)^{2}}{b c} \geq \frac{4 b c}{b c}=4
$$

is an obvious inequality - where in the second step, we used the relationship between the arithmetic and geometric means - from which the left-hand side inequality follows.

According to the conditions, if $0<c$, the following is valid: $4 b-c>2 b-c \geq a+b-c>0$.

Therefore, $(4 b-c)(b-c) \leq 0$, and thus from (2) $\frac{(a+b+c)^{2}}{b c} \leq 9$. This proves (1).

We will show that the expression under consideration takes on every value greater than 4 but not greater than 9. Let $b=c=1$, and consider the expression $f(a)=\frac{(a+b+c)^{2}}{b c}=(a+2)^{2}$. If $0<a \leq 1$, then $a \leq b \leq c$, and the triangle inequalities are also satisfied. The function $f(a)=(a+2)^{2}$ is continuous on the interval $[0 ; 1]$, strictly monotonic, $f(0)=4, f(1)=9$, so this function takes on every value greater than 4 and not greater than 9 on the interval $] 0 ; 1]$. Therefore, the range of the expression $\frac{(a+b+c)^{2}}{b c}$ is the interval $\left.] 4 ; 9\right]$.