If $a^2 + b^2 + c^2 = 1$, what is the maximum value of $abc$?

### Concept & Formula This problem is solved using the **AM-GM Inequality** (Arithmetic Mean - Geometric Mean Inequality). For any set of non-negative real numbers, the arithmetic mean is always greater than or equal to the geometric mean. $$\frac{x + y + z}{3} \ge \sqrt[3]{xyz}$$ ### Step-by-Step Solution * **Given:** $a^2 + b^2 + c^2 = 1$ * **Calculation:** Let's apply the AM-GM inequality to the three non-negative terms: $a^2$, $b^2$, and $c^2$. $$\frac{a^2 + b^2 + c^2}{3} \ge \sqrt[3]{a^2 \times b^2 \times c^2}$$ * Substitute the given value ($a^2 + b^2 + c^2 = 1$) into the left side of the inequality: $$\frac{1}{3} \ge \sqrt[3]{(abc)^2}$$ * To isolate $abc$, first cube both sides to remove the cube root: $$\left(\frac{1}{3}\right)^3 \ge (abc)^2$$ $$\frac{1}{27} \ge (abc)^2$$ * Now, take the square root of both sides. Since we are looking for the maximum positive value, we take the positive root: $$\sqrt{\frac{1}{27}} \ge abc$$ $$\frac{1}{\sqrt{9 \times 3}} \ge abc$$ $$\frac{1}{3\sqrt{3}} \ge abc$$ * Therefore, the maximum possible value of $abc$ is $\frac{1}{3\sqrt{3}}$. ### Exam Strategy & Shortcut In optimization problems involving squares or symmetric conditions (like $a^2 + b^2 + c^2 = 1$), the maximum or minimum product almost always occurs when all variables are equal. Assume **Symmetry**: $a = b = c$. Substitute this into the given equation: $3a^2 = 1 \implies a^2 = 1/3 \implies a = 1/\sqrt{3}$. The maximum product $abc$ is simply $a \times a \times a = a^3$. Calculate: $(1/\sqrt{3})^3 = 1 / (3\sqrt{3})$. You arrive at the answer in seconds without formal inequalities. ### Common Pitfall Students often try to use calculus (Lagrange multipliers) or trigonometry (spherical coordinates) to solve this. While technically correct, those methods are far too time-consuming for an aptitude exam. Always check for symmetry or AM-GM first. ### Final Answer Therefore, the correct answer is **$\frac{1}{3\sqrt{3}}$**.