Solving a Romanian Olympiad Problem: Sum of Square Roots Without a Calculator

The Challenge: Summing Square Roots

• 🎯 The problem asks to prove that the sum of square roots from $\sqrt{5}$ to $\sqrt{13}$ (inclusive) is between 26 and 27, without using a calculator.
• 💡 This problem, originating from the Romanian Mathematical Olympiad, is designed to test mathematical ingenuity, even for talented young students.
• 🔢 The sum, denoted by 'n', includes nine terms: $\sqrt{5} + \sqrt{6} + \sqrt{7} + \sqrt{8} + \sqrt{9} + \sqrt{10} + \sqrt{11} + \sqrt{12} + \sqrt{13}$.

Method 1: Completing the Square and Pairing

• 🧩 This method involves pairing terms symmetrically (e.g., $\sqrt{5} + \sqrt{13}$, $\sqrt{6} + \sqrt{12}$) and using geometric constructions to estimate the sum of each pair.
• 📐 By completing the square with areas related to the radicands, the sum of a pair like $\sqrt{5} + \sqrt{13}$ can be bounded using perfect squares (e.g., $\sqrt{18 + 2\sqrt{65}}$).
• 📈 An upper bound of 6 and a lower bound of 5.75 were established for each pair by strategically choosing values close to $\sqrt{65}$ (like $\sqrt{81}$ and $\sqrt{64}$).
• ✅ Summing the bounds for the four pairs and the middle term $\sqrt{9}=3$ confirms that $4 \times 5.75 + 3 < n < 4 \times 6 + 3$, thus $26 < n < 27$.

Method 2: Using a Concavity Inequality

• 💡 This method utilizes the inequality $\sqrt{a} + \sqrt{b} > \sqrt{a-x} + \sqrt{b+x}$ for $b \ge a \ge x > 0$, derived from the concave nature of the square root function.
• 📉 Graphically, this inequality shows that moving 'x' units inward from 'a' decreases the square root value more than moving 'x' units outward from 'b' increases it.
• 🚀 Applying this inequality to pairs like $\sqrt{5} + \sqrt{8}$ with $x=1$ yields a lower bound of 5, and to pairs like $\sqrt{5} + \sqrt{13}$ with $x=4$ yields an upper bound of 6.
• ✨ The algebraic proof confirms the inequality by comparing squared terms and simplifying to $0 \ge x(a-b) - x^2$, which holds true given the constraints.

Method 3: Trapezoid Areas and Jensen's Inequality

• 📊 Jensen's inequality for concave functions states that $\frac{\sqrt{a} + \sqrt{b}}{2} \le \sqrt{\frac{a+b}{2}}$, implying $\sqrt{a} + \sqrt{b} \le 2\sqrt{\frac{a+b}{2}}$.
• 📈 This inequality is used to establish the upper bound: each pair $\sqrt{a} + \sqrt{b}$ is less than $2\sqrt{\text{average}(a,b)}$, leading to an upper bound of 6 for each pair and $n < 27$.
• 📐 For the lower bound, the area under the square root curve is approximated by summing trapezoid areas between consecutive integers.
• 🔗 The sum of these trapezoid areas provides a lower bound for the integral of $\sqrt{x}$, which, after manipulation, leads to $n > 4.5 \times (\sqrt{5} + \sqrt{13})$, and using $\sqrt{34} > 5.8$, confirms $n > 26.1$.

Method 4: Inequality of Means

• ⚖️ This method employs the inequality of means: Harmonic Mean $\le$ Geometric Mean $\le$ Arithmetic Mean $\le$ Quadratic Mean.
• 🚀 The Quadratic Mean (QM) inequality ($\frac{\sum x_i}{n} \le \sqrt{\frac{\sum x_i^2}{n}}$) is used to establish the upper bound: the average of the square roots is less than the square root of the average of the radicands, leading to $n < 27$.
• 🔗 The Arithmetic Mean-Geometric Mean (AM-GM) inequality ($\frac{a+b}{2} \ge \sqrt{ab}$) is adapted to show $\sqrt{a} + \sqrt{b} > 2\sqrt{\sqrt{ab}}$.
• 🔑 By carefully selecting pairs and estimating fourth roots (e.g., $\sqrt[4]{81}=3$, $\sqrt[4]{65} \approx 2.8$, $\sqrt[4]{60} \approx 2.78$, $\sqrt[4]{54} \approx 2.71$), lower bounds for each pair are found, summing to $n > 26$.