0
$\begingroup$

Let $x_1, x_2, x_3$ be real numbers. First show $$ (x_1+ x_2)^2 + (x_1+ x_3)^2 + (x_2+ x_3)^2 ≥ x_1^2 + x_2^2 + x_3^2, $$, then use Cauchy-Schwarz inequality to show it.

My try:

Let $u^{\top}=[x_1+x_2, x_1+x_3, x_2+x_3]^{\top}$ and $v^{\top}=[x_1, x_2, x_3]^{\top}$ or any other combinations that makes sense.

$\endgroup$
5
  • $\begingroup$ From Cauchy-Schwarz inequality we have: $(a^2+b^2+c^2)(1+1+1)≥(a+b+c)^2$ $|a=x_1+x_2|b=x_1+x_3|c=x_2+x_3|$ $\implies$ $\left(2\left(x_1+x_2+x_3\right)\right)^2 \ge 3({x_1}^2+{x_2}^2+{x_3}^2)$ $\endgroup$
    – Rehman
    Feb 9, 2023 at 13:52
  • $\begingroup$ @Rehman: Can you please add more details on how you get 3 and 2 in your second comment. Also, please post your comments as solutions. $\endgroup$
    – Saeed
    Feb 9, 2023 at 14:00
  • $\begingroup$ Of course :) If you find any mistake in my post, I would appreciate it if you let me know $\endgroup$
    – Rehman
    Feb 9, 2023 at 14:26
  • $\begingroup$ The identity $(a+b)^2+(b+c)^2+(c+a)^2=a^2+b^2+c^2+(a+b+c)^2$ is a self evident and simple proof for the inequality. Why would you want to use CS inequality specifically instead to solve this one? $\endgroup$
    – Macavity
    Feb 10, 2023 at 13:03
  • $\begingroup$ @Macavity: I want to expand it for $x_1,\dots,x_5$ where the left hand side is sum of three-term component and right is sum of all the two-term components. $\endgroup$
    – Saeed
    Feb 10, 2023 at 18:05

1 Answer 1

1
$\begingroup$

$$(x_1+x_2+x_3)^2 \ge 0$$ $$\left({x_1}^2+{x_2}^2+{x_3}^2\right) + 2(x_1x_2+x_1x_3+x_2x_3) \ge 0 $$ $$2\left({x_1}^2+{x_2}^2+{x_3}^2\right) + 2(x_1x_2+x_1x_3+x_2x_3)\ge {x_1}^2+{x_2}^2+{x_3}^2$$ $$ (x_1+x_2)^2+(x_1+x_3)^2+(x_2+x_3)^2=2\left({x_1}^2+{x_2}^2+{x_3}^2\right) + 2(x_1x_2+x_1x_3+x_2x_3)$$ $$(x_1+x_2)^2+(x_1+x_3)^2+(x_2+x_3)^2 \ge {x_1}^2+{x_2}^2+{x_3}^2 $$

$\endgroup$
5
  • 1
    $\begingroup$ I cannot see how from $${x_1}^2+{x_2}^2+{x_3}^2+ 8(x_1x_2+x_1x_3+x_2x_3) \ge 0$$ you get $$x_1x_2+x_1x_3+x_2x_3 \ge 3\sqrt[3]{(x_1x_2x_3)^2} .$$ It is not clear how you are a.m-g.m. Can you fill the gaps here. $\endgroup$
    – Saeed
    Feb 9, 2023 at 14:44
  • $\begingroup$ @Sepide A.M-G.M inequality: $$ x+y+z \ge 3\sqrt[3]{xyz}$$ $|x=x_1x_2|y=x_1x_3|z=x_2x_3|$ $$x_1x_2+x_1x_3+x_2x_3 \ge 3\sqrt[3]{x_1x_2 \cdot x_1x_3 \cdot x_2x_3} $$ $ x_1x_2 \cdot x_1x_3 \cdot x_2x_3= (x_1x_2x_3)^2$ $$x_1x_2+x_1x_3+x_2x_3 \ge 3\sqrt[3]{(x_1x_2x_3)^2}$$ $\endgroup$
    – Rehman
    Feb 9, 2023 at 15:08
  • $\begingroup$ @Macavity Ohhh sorry..... I completely forgot about that $\endgroup$
    – Rehman
    Feb 9, 2023 at 15:21
  • $\begingroup$ @Sepide What Martin said is true. Therefore the last line (if true) must be proved in a different way. $\endgroup$
    – Rehman
    Feb 9, 2023 at 15:24
  • $\begingroup$ @Macavity Yes, you are right. I tried a few more times, but each time it made the right side of the inequality too small. What do you think is the reason for this? Do you think this is because the Cauchy-Schwarz inequality reduces the inequality too much? Am I just not applying the inequality correctly? Or what do you think? $\endgroup$
    – Rehman
    Feb 9, 2023 at 16:13

Not the answer you're looking for? Browse other questions tagged .