2026/006) Prove that for real $x_1,x_2,x_3$ $x_1^2 + x_2^2+ x_3^2 \ge \frac{4}{3}(x_1x_2 + x_2x_3)$

We have

 $x_1^2 + x_2^2+ x_3^2-\frac{4}{3}(x_1x_2 + x_2x_3)$

$=(x_1^2- \frac{4}{3}x_1x_2) + ( x_3^2-  \frac{4}{3} x_2x_3)+x_2^2$  Combining $x_1x_2$ with $x_1^2$ term and   $x_3x_2$ with $x_2^2$ term 

$=(x_1^2- 2 *\frac{2}{3}x_1x_2 + (\frac{2}{3}x_2)^2) + ( x_3^2-  2* ( \frac{2}{3} x_2x_3+  (\frac{2}{3}x_2)^2)+(x_2^2-  (\frac{2}{3}x_2)^2) - (\frac{2}{3}x_2)^2)$  completing the square and subtracting the same 

$=(x_1 - \frac{2}{3}x_2)^2 + ( x_3- \frac{2}{3}x_2)^2+\frac{1}{9} x_2^2$

This is positive or zero

So 

  $x_1^2 + x_2^2+ x_3^2-\frac{4}{3}(x_1x_2 + x_2x_3)>=  0$

Hence  $x_1^2 + x_2^2+ x_3^2 \ge \frac{4}{3}(x_1x_2 + x_2x_3)$