# Application of Cauchy-Schwarz inequality for 3 real numbers [closed]

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.

• From Cauchy-Schwarz inequality we have: $(a^2+b^2+c^2)(1+1+1)≥(a+b+c)^2$ $|a=x_1+x_2|b=x_1+x_3|c=x_2+x_3|$ $\implies$ $\left(2\left(x_1+x_2+x_3\right)\right)^2 \ge 3({x_1}^2+{x_2}^2+{x_3}^2)$ Feb 9, 2023 at 13:52
• 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? Feb 10, 2023 at 13:03
• @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. Feb 10, 2023 at 18:05
$$(x_1+x_2+x_3)^2 \ge 0$$ $$\left({x_1}^2+{x_2}^2+{x_3}^2\right) + 2(x_1x_2+x_1x_3+x_2x_3) \ge 0$$ $$2\left({x_1}^2+{x_2}^2+{x_3}^2\right) + 2(x_1x_2+x_1x_3+x_2x_3)\ge {x_1}^2+{x_2}^2+{x_3}^2$$ $$(x_1+x_2)^2+(x_1+x_3)^2+(x_2+x_3)^2=2\left({x_1}^2+{x_2}^2+{x_3}^2\right) + 2(x_1x_2+x_1x_3+x_2x_3)$$ $$(x_1+x_2)^2+(x_1+x_3)^2+(x_2+x_3)^2 \ge {x_1}^2+{x_2}^2+{x_3}^2$$
• 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. Feb 9, 2023 at 14:44
• @Sepide A.M-G.M inequality: $$x+y+z \ge 3\sqrt[3]{xyz}$$ $|x=x_1x_2|y=x_1x_3|z=x_2x_3|$ $$x_1x_2+x_1x_3+x_2x_3 \ge 3\sqrt[3]{x_1x_2 \cdot x_1x_3 \cdot x_2x_3}$$ $x_1x_2 \cdot x_1x_3 \cdot x_2x_3= (x_1x_2x_3)^2$ $$x_1x_2+x_1x_3+x_2x_3 \ge 3\sqrt[3]{(x_1x_2x_3)^2}$$ Feb 9, 2023 at 15:08