Balti Tee 2021 · Valikvooru ülesanne

Given are three real numbers $a, b, c$ such that $0 < a, b, c < \frac{a+b+c}{2}$ (the inequality $a, b, c < \frac{a+b+c}{2}$ is equivalent to the triangle inequality). Introduce the variables

$$x = \frac{-a + b + c}{2(a + b + c)}, \quad y = \frac{a - b + c}{2(a + b + c)} \quad \text{and} \quad z = \frac{a + b - c}{2(a + b + c)}$$

The conditions imposed on $a, b$ and $c$ are equivalent to

$$0 < x, y, z, \quad \text{and} \quad x + y + z = \frac{1}{2}$$

Clearly

$$\frac{b+c}{a+b+c} = \frac{1}{2} + x, \quad \frac{c+a}{a+b+c} = \frac{1}{2} + y \quad \text{and} \quad \frac{a+b}{a+b+c} = \frac{1}{2} + z$$

Therefore we have:

$$\begin{align*} \frac{(a+b)(a+c)(b+c)}{(a+b+c)^3} &= \frac{b+c}{a+b+c} \cdot \frac{c+a}{a+b+c} \cdot \frac{a+b}{a+b+c} \\ &= \left(\frac{1}{2} + x\right) \cdot \left(\frac{1}{2} + y\right) \cdot \left(\frac{1}{2} + z\right) \\ &= \frac{1}{8} + \frac{x+y+z}{4} + \frac{yz+zx+xy}{2} + xyz \\ &= \frac{1}{8} + \frac{1/2}{4} + \frac{yz+zx+xy}{2} + xyz \\ &> \frac{1}{4} \end{align*}$$

Without loss of generality it may be assumed that $a \le b \le c$. Observe that

$$((a-t)+b)(b+(c+t)) = (a+b)(b+c) - t(c-a) - t^2 < (a+b)(b+c).$$

for $t > 0$. The initial numbers $a, b, c$ satisfy the triangle inequality $a + b > c$. Choose $t$ such that

$$(a-t)+b=c+t,$$

it is clear that $0 < t < a$. If we replace in the inequality under consideration $a$ and $c$ with $a-t, c+t$, the sums $a+c$ and $a+b+c$ remain unchanged, and the product $(a+b)(b+c)$ decreases. Hence it is sufficient to prove the given inequality for positive $a, b, c$ under assumption $a+b=c$. In this case

$$4 \cdot \frac{a+b}{a+b+c} \cdot \frac{a+c}{2c} \cdot \frac{b+c}{2c} = 4 \cdot \frac{1}{2} \cdot \frac{a+c}{2c} \cdot \frac{b+c}{2c} = \frac{ab+c(a+b)+c^2}{2c^2} = \frac{ab+2c^2}{2c^2} > 1.$$