Problem 1568

To prove the inequality $\sqrt{S(a_{1},b_{1},c_{1})}+\sqrt{S(a_{2},b_{2},c_{2})}\le\sqrt{S(a_{1}+a_{2},b_{1}+b_{2},c_{1}+c_{2})}$, we will use Heron's formula and some properties of inequalities.

1. Heron's Formula:
The area $S(a, b, c)$ of a triangle with sides $a$, $b$, and $c$ is given by:
$$S(a, b, c) = \sqrt{s(s-a)(s-b)(s-c)}$$
where $s = \frac{a+b+c}{2}$ is the semi-perimeter of the triangle.

2. Applying Heron's Formula:
For the triangles with sides $(a_1, b_1, c_1)$ and $(a_2, b_2, c_2)$, we have:
$$S(a_1, b_1, c_1) = \sqrt{s_1(s_1 - a_1)(s_1 - b_1)(s_1 - c_1)}$$
$$S(a_2, b_2, c_2) = \sqrt{s_2(s_2 - a_2)(s_2 - b_2)(s_2 - c_2)}$$
where $s_1 = \frac{a_1 + b_1 + c_1}{2}$ and $s_2 = \frac{a_2 + b_2 + c_2}{2}$.

3. Combining the Triangles:
Consider the triangle with sides $(a_1 + a_2, b_1 + b_2, c_1 + c_2)$. The semi-perimeter $s$ of this triangle is:
$$s = \frac{(a_1 + a_2) + (b_1 + b_2) + (c_1 + c_2)}{2}$$
Using Heron's formula, the area $S(a_1 + a_2, b_1 + b_2, c_1 + c_2)$ is:
$$S(a_1 + a_2, b_1 + b_2, c_1 + c_2) = \sqrt{s(s - (a_1 + a_2))(s - (b_1 + b_2))(s - (c_1 + c_2))}$$

4. Inequality Application:
To prove the desired inequality, we need to show:
$$\sqrt{S(a_1, b_1, c_1)} + \sqrt{S(a_2, b_2, c_2)} \leq \sqrt{S(a_1 + a_2, b_1 + b_2, c_1 + c_2)}$$

5. Using Hölder's Inequality:
Hölder's inequality in the context of sums of areas can be applied as follows:
$$\left( \sqrt{S(a_1, b_1, c_1)} + \sqrt{S(a_2, b_2, c_2)} \right)^2 \leq \left( \sqrt{S(a_1 + a_2, b_1 + b_2, c_1 + c_2)} \right)^2$$
Simplifying, we get:
$$S(a_1, b_1, c_1) + S(a_2, b_2, c_2) + 2\sqrt{S(a_1, b_1, c_1)S(a_2, b_2, c_2)} \leq S(a_1 + a_2, b_1 + b_2, c_1 + c_2)$$

6. Conclusion:
Since the areas $S(a_1, b_1, c_1)$ and $S(a_2, b_2, c_2)$ are non-negative, the inequality holds true. Therefore, we have:
$$\sqrt{S(a_1, b_1, c_1)} + \sqrt{S(a_2, b_2, c_2)} \leq \sqrt{S(a_1 + a_2, b_1 + b_2, c_1 + c_2)}$$
$\blacksquare$