Problem 1358

$$\frac{s_{a}^{2}}{b c}+\frac{s_{a}^{2}}{c a}+\frac{s_{a}^{2}}{a b} \geq \frac{9}{4}$$

I. Solution. The medians can be expressed in terms of the sides. Reflecting the triangle over the midpoint of side $a$, the sides of the resulting parallelogram are $b$ and $c$, and the diagonals are $a$ and $2 s_{a}$. By repeatedly applying the Pythagorean theorem, it is easy to see that the sum of the squares of the diagonals of a parallelogram is equal to the sum of the squares of its sides, so in our case

$$a^{2}+4 s_{a}^{2}=2\left(b^{2}+c^{2}\right), \quad s_{a}^{2}=\left(2 b^{2}+2 c^{2}-a^{2}\right) / 4$$

Substituting this and similar expressions for $s_{b}^{2}$ and $s_{c}^{2}$ into (1), multiplying by $4 a b c$, and rearranging, the resulting inequality

$$a\left(2 b^{2}+2 c^{2}-a^{2}\right)+b\left(2 c^{2}+2 a^{2}-b^{2}\right)+c\left(2 a^{2}+2 b^{2}-c^{2}\right)-9 a b c \geq 0$$

is equivalent to the statement.

Further transforming the left side,

$$2 a(b-c)^{2}+2 b(c-a)^{2}+2 c(a-b)^{2}-\left(a^{3}+b^{3}+c^{3}-3 a b c\right) \geq 0$$

and since the subtracted four-term expression can be written as:

$$\frac{1}{2}(a+b+c)\left\{(a-b)^{2}+(b-c)^{2}+(c-a)^{2}\right\}$$

it suffices to prove:

$$\begin{gathered} (b-c)^{2}\left(2 a-\frac{a+b+c}{2}\right)+(c-a)^{2}\left(2 b-\frac{a+b+c}{2}\right)+ \\ +(a-b)^{2}\left(2 c-\frac{a+b+c}{2}\right) \geq 0 \end{gathered}$$

Based on the symmetry of the expression, we can choose the labeling of the triangle's sides such that $a \geq b \geq c(>0)$. Then the first two products on the left side of (2) are non-negative, since

$$\begin{gathered} 2 a-\frac{a+b+c}{2}=\frac{1}{2}(a+(a-b)+(a-c))>0 \\ 2 b-\frac{a+b+c}{2}=\frac{1}{2}(2 b+b-a-c) \geq \frac{1}{2}(2 c+b-a-c)=\frac{1}{2}(b+c-a)>0 \end{gathered}$$

The third product, $P_{3}$, could be negative; however, we will show that the sum of the second product $P_{2}$ and $P_{3}$ is non-negative. Indeed, substituting $(b-a)^{2}$ for $(c-a)^{2}$ in $P_{2}$ does not increase it according to (3):

$$P_{2}+P_{3} \geq(b-a)^{2}\left(2 b-\frac{a+b+c}{2}\right)+(a-b)^{2}\left(2 c-\frac{a+b+c}{2}\right)=(a-b)^{2}(b+c-a)$$

from which our statement is evident, and thus we have proven (1) according to the preceding.

II. Solution. From the solution of problem 1856, we can read as an intermediate result that if $A, B, C$ are the vertices of a triangle, its sides are $A B=c, B C=a$, and $C A=b$, $P$ is any point in space, and $\alpha, \beta, \gamma$ are positive numbers, then

$$\alpha \cdot P A^{2}+\beta \cdot P B^{2}+\gamma \cdot P C^{2} \geq \frac{\alpha \beta c^{2}+\beta \gamma a^{2}+\gamma \alpha b^{2}}{\alpha+\beta+\gamma}$$

Let the weights now be

$$\alpha=\frac{1}{b c}, \quad \beta=\frac{1}{c a}, \quad \gamma=\frac{1}{a b}$$

and specifically, taking $P$ as the centroid of triangle $A B C$,

$$P A=\frac{2}{3} s_{a}, \quad P B=\frac{2}{3} s_{b}, \quad P C=\frac{2}{3} s_{c}$$

and substituting these into the problem statement, we obtain the result.[^0]

[^0]: ${ }^{1}$ See in this issue, on page 122, (5), where the specific values of $\alpha, \beta$ and $\gamma$ have not yet been considered.