Solution: Use the Cauchy-Schwarz inequality: $(2^2 + 3^2 + 4^2)(x^2 + y^2 + z^2) \geq (2x + 3y + 4z)^2$. This gives $29(x^2 + y^2 + z^2) \geq 144$, so $x^2 + y^2 + z^2 \geq rac{144}{29}$. Equality holds when $rac{x}{2} = rac{y}{3} = rac{z}{4} = k$, leading to $x = 2k$, $y = 3k$, $z = 4k$. Substituting into $2x + 3y + 4z = 12$ gives $4k + 9k + 16k = 29k = 12$, so $k = rac{12}{29}$. Thus, the minimum value is $oxed{\dfrac{144}{29}}$.

Sat, 02 May 2026 20:46:18 +0000

Content under review