Question

Problem 6.1. Let x1, . . . , xn ∈ R and let a1, . . . , an > 0. a) Show that a1x 2 1 + . . . + anx 2 n ≥ 0. b) Show that a1x 2 1 + . . . + anx 2 n = 0 if and only if x1 = · · · = xn = 0

Answer 1

To show that a1x1^2 + ... + anx2n is always non-negative, we note that the square of any real number is non-negative and that the product of a positive number with a non-negative number is non-negative. To prove that x1, ..., xn must be zero if a1x1^2 + ... + anx2n equals zero, we argue that since ai is positive, each xi must be zero because only the square of zero is zero.

Step-by-step explanation:

The student's question can be addressed by recognizing that squares of real numbers are always non-negative and that a product of a non-negative number with a positive number is also non-negative.

Part a)

The expression a1x1^2 + ... + anx2n consists of terms ai xi^2, where ai > 0 for all i, and xi is any real number. The square of any real number xi^2 is non-negative, and since ai is positive, their product is non-negative. Since a sum of non-negative numbers is non-negative, the entire expression is greater than or equal to zero.

Part b)

For the equation a1x1^2 + ... + anx2n = 0 to hold, each individual term ai xi^2 must be zero because adding positive numbers cannot result in zero. Since ai > 0, it requires that each xi^2 must be zero, and the only real number whose square is zero is zero itself. Thus, x1, ..., xn must all equal zero.