Let V be the volume of the parallelopiped formed by the vectors a = a1i + a2j + a3k, b = b1i + b2j +

Q: Why can we use the AM-GM inequality here?

The AM-GM inequality is applicable because the problem explicitly states that $a_r, b_r, c_r$ are non-negative real numbers. This ensures that the sums $S_1, S_2, S_3$ are also non-negative, which is the fundamental requirement for applying AM-GM.

Q: Why is the volume $V$ less than or equal to the product $S_1 S_2 S_3$?

The product $S_1 S_2 S_3$ expands into 27 non-negative terms, which include the three positive terms of the determinant $V$. Since $V$ is equal to the sum of these three positive terms minus three other non-negative terms, $V$ must be less than or equal to the sum of the three positive terms, which in turn is less than or equal to the sum of all 27 non-negative terms (i.e., $S_1 S_2 S_3$).

Q: Is it possible for the volume to actually equal $L^3$?

Yes, the equality $V = L^3$ holds when the conditions for equality in AM-GM are met, which means $S_1 = S_2 = S_3 = L$. Additionally, for the volume to reach this maximum, the negative terms in the determinant expansion must be zero, which occurs when the vectors are orthogonal and aligned with the axes.

Question

Let $V$ be the volume of the parallelopiped formed by the vectors $\vec{a} = a_1\hat{i} + a_2\hat{j} + a_3\hat{k}$, $\vec{b} = b_1\hat{i} + b_2\hat{j} + b_3\hat{k}$, $\vec{c} = c_1\hat{i} + c_2\hat{j} + c_3\hat{k}$. If $a_r, b_r, c_r$, where $r = 1, 2, 3$, are non-negative real numbers and $\sum_{r=1}^3 (a_r + b_r + c_r) = 3L$, show that $V \le L^3$.

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine you are standing in a 3D coordinate system, holding three vectors $\vec{a}, \vec{b}, \vec{c}$ that originate from the origin. These vectors are the edges of a parallelepiped, a slanted box that captures the essence of 3D space.

Our mission is to find the maximum possible volume $V$ of this box, given a constraint on the sum of its components. This is not just an algebraic exercise; it is a journey into the heart of geometric optimization.

### The Volume as a Determinant

The volume $V$ of a parallelepiped is defined by the scalar triple product of its edge vectors. Mathematically, this is the absolute value of the determinant of a $3 	imes 3$ matrix:

$$V = \left| \begin{vmatrix} a_1 & a_2 & a_3 \ b_1 & b_2 & b_3 \ c_1 & c_2 & c_3 \end{vmatrix} ight|$$

When we expand this determinant, we get six terms: three positive and three negative. The positive terms are $a_1b_2c_3 + a_2b_3c_1 + a_3b_1c_2$, and the negative terms are $a_1b_3c_2 + a_2b_1c_3 + a_3b_2c_1$.

This structure is the key to our proof.

### The Constraint and the Power of AM-GM

We are given the constraint $\sum_{r=1}^3 (a_r + b_r + c_r) = 3L$. To simplify this, let us define the sum of components along each axis: $S_1 = a_1 + b_1 + c_1$, $S_2 = a_2 + b_2 + c_2$, and $S_3 = a_3 + b_3 + c_3$.

The constraint then becomes $S_1 + S_2 + S_3 = 3L$. Since all components are non-negative, $S_1, S_2, S_3$ are also non-negative.

This is the perfect setup for the Arithmetic Mean-Geometric Mean (AM-GM) inequality:

$$\frac{S_1 + S_2 + S_3}{3} \ge (S_1 S_2 S_3)^{1/3}$$

Substituting our constraint, we get $\frac{3L}{3} \ge (S_1 S_2 S_3)^{1/3}$, which simplifies to $L \ge (S_1 S_2 S_3)^{1/3}$. Cubing both sides, we arrive at the powerful inequality:

$$L^3 \ge S_1 S_2 S_3$$

### The Final Connection

Now, let us look at the product $S_1 S_2 S_3 = (a_1+b_1+c_1)(a_2+b_2+c_2)(a_3+b_3+c_3)$. Expanding this product generates 27 terms. Because all $a_r, b_r, c_r$ are non-negative, all 27 terms are non-negative.

Crucially, the three positive terms of our volume determinant ($a_1b_2c_3, a_2b_3c_1, a_3b_1c_2$) are among these 27 terms. Therefore, $S_1 S_2 S_3 \ge a_1b_2c_3 + a_2b_3c_1 + a_3b_1c_2$.

Since $V$ is equal to these three positive terms minus the three negative terms, and the negative terms are also non-negative, it follows that $V \le a_1b_2c_3 + a_2b_3c_1 + a_3b_1c_2$. By transitivity, we have $L^3 \ge S_1 S_2 S_3 \ge V$.

Thus, we have shown that **$V \le L^3$**. This result is a beautiful testament to how inequalities can bound geometric shapes, turning a complex 3D problem into a simple, elegant conclusion.

Visualized Solution

The Parallelepiped

Volume as Scalar Triple Product

Grouping by Components

Total Sum Condition

AM-GM Inequality

Substituting the Sum

Cubing Both Sides

Expanding $S_{1} S_{2} S_{3}$

Expanding the Volume Determinant

Bounding the Volume

Final Inequality

The Sigma Insight: Scalar Triple Product

Analyzing the Setup

The Volume as a Determinant

The Constraint and the Power of AM-GM

The Final Connection

Similar Questions

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Match List I with List II:

List-I

List-II

If $a = α \hat{i} + β \hat{j} + 3 \hat{k}$ , $b = β \hat{i} - α \hat{j} - \hat{k}$ and $c = \hat{i} - 2 \hat{j} - \hat{k}$ such that $a \cdot b = 1$ and $b \cdot c = - 3$ , then $\frac{1}{3} ((a \times b) \cdot c)$ is equal to

Let $a, b, c$ be three non-coplanar vectors and $p, q, r$ are vectors defined by the relations $p = \frac{b \times c}{[ a b c ]}, q = \frac{c \times a}{[ a b c ]}, r = \frac{a \times b}{[ a b c ]}$ then the value of the expression $(a + b) \cdot p + (b + c) \cdot q + (c + a) \cdot r$ is equal to