Question

The least value of $x_1^2+x_2^2+x_3^2$ where $x_1,x_2,x_3$ are real numbers satisfying $x_1+2x_2+3x_3=4$ is

• A. $(\frac{8}{7}{)}^{\frac{5}{3}}$
• B. $1$
• C. $4$
• D. $2$

Answer 1

The correct option is A $(\frac{8}{7}{)}^{\frac{5}{3}}$

$\frac{x_1+2x_2+3x_3}{3}\ge {\left[\right(x_1{x}_2{x}_3\left)6\right]}^{1/3}\frac{1}{6}{\left(4/3\right)}^3\ge \left(x_1{x}_2{x}_3\right)\left(1\right)\frac{{x_1}^2+{x_2}^2+{x_3}^2}{3}\ge {\left(x_1{x}_2{x}_3\right)}^{2/3}{\left[\frac{{x_1}^2+{x_2}^2+{x_3}^2}{3}\right]}^{3/2}\ge \left(x_1{x}_2{x}_3\right)\left(2\right)$

Dividing $\left(2\right)$ by $\left(1\right)$

${\left[\frac{{x_1}^2+{x_2}^2+{x_3}^2}{3}\right]}^{3/2}\ge \frac{1}{6}{\left(4/3\right)}^3{[{x_1}^2+{x_2}^2+{x_3}^2]}^{3/2}\ge \frac{3^{3/2}{2}^5}{3^4}{x_1}^2+{x_2}^2+{x_3}^2\ge {\left[\frac{2^5}{3^{5/2}}\right]}^{2/3}{x_1}^2+{x_2}^2+{x_3}^2\ge \frac{2^{10/3}}{3^{5/3}}{x_1}^2+{x_2}^2+{x_3}^2\ge {\left(\frac{4}{3}\right)}^{5/3}$