Can we prove that $|a\parallel b\parallel c|\geq (|a|+|b|+|c|)^{3}$ with $a+b+c=0?$ If we can, I could prove the inequality in question.

Do you have any typos ? Since $(|a|+|b|+|c|)^{3}\geq 27 |a\parallel b\parallel c|$ ???

Anh Cuong wrote: Do you have any typos ? Since $(|a|+|b|+|c|)^{3}\geq 27 |a\parallel b\parallel c|$ ??? Hum, you are right. Here is my solution. Let $x=\frac{|c|}{|a|+|b|+|c|},\ y=\frac{|a|}{|a|+|b|+|c|},\ z=\frac{|b|}{|a|+|b|+|c|},$ we have $\frac{|a|+|b|+(2\lambda+1)|c|}{|a|+|b|+|c|}=1+\frac{2\lambda |c|}{|b|+|c|+|a|}=1+2\lambda x,$ and $x+y+z=1,\ x\geq 0,\ y\geq 0,\ z\geq 0.$ simillarly $\frac{|b|+|c|+(2\lambda+1)|a|}{|a|+|b|+|c|}=1+2\lambda y,\ \frac{|c|+|a|+(2\lambda+1)|b|}{|c|+|a|+|b|}=1+2\lambda z.$ Thus $\frac{|a|+|b|+(2\lambda+1)|c|}{|a|+|b|+|c|}\cdot \frac{|b|+|c|+(2\lambda+1)|a|}{|a|+|b|+|c|}\cdot \frac{|c|+|a|+(2\lambda+1)|b|}{|c|+|a|+|b|}$ $=(1+2\lambda x)(1+2\lambda y)(1+2\lambda z)$ $=1+2\lambda (x+y+z)+4\lambda^{2}(xy+yz+zx)+8\lambda^{3}xyz$ $=(2\lambda+1)(x+y+z)^{3}+4\lambda^{2}(x+y+z)(xy+yz+zx)+8\lambda^{3}xyz$ $\geq (2\lambda+1)(3\sqrt[3]{xyz})^{3}+4\lambda^{2}\cdot 3\sqrt[3]{xyz}\cdot 3\sqrt[3]{(xyz)^{2}}+8\lambda^{3}xyz$ $=\left\{27(2\lambda+1)+36\lambda^{2}+8\lambda^{3}\right\}xyz$ $=(8\lambda^{3}+36\lambda^{2}+54\lambda+27)xyz$ $=(2\lambda+3)^{3}xyz$

Kunihiko_Chikaya wrote: Anh Cuong wrote: Do you have any typos ? Since $(|a|+|b|+|c|)^{3}\geq 27 |a\parallel b\parallel c|$ ??? Hum, you are right. Here is my solution. Let $x=\frac{|c|}{|a|+|b|+|c|},\ y=\frac{|a|}{|a|+|b|+|c|},\ z=\frac{|b|}{|a|+|b|+|c|},$ we have $\frac{|a|+|b|+(2\lambda+1)|c|}{|a|+|b|+|c|}=1+\frac{2\lambda |c|}{|b|+|c|+|a|}=1+2\lambda x,$ and $x+y+z=1,\ x\geq 0,\ y\geq 0,\ z\geq 0.$ simillarly $\frac{|b|+|c|+(2\lambda+1)|a|}{|a|+|b|+|c|}=1+2\lambda y,\ \frac{|c|+|a|+(2\lambda+1)|b|}{|c|+|a|+|b|}=1+2\lambda z.$ Thus $\frac{|a|+|b|+(2\lambda+1)|c|}{|a|+|b|+|c|}\cdot \frac{|b|+|c|+(2\lambda+1)|a|}{|a|+|b|+|c|}\cdot \frac{|c|+|a|+(2\lambda+1)|b|}{|c|+|a|+|b|}$ $=(1+2\lambda x)(1+2\lambda y)(1+2\lambda z)$ $=1+2\lambda (x+y+z)+4\lambda^{2}(xy+yz+zx)+8\lambda^{3}xyz$ $=(2\lambda+1)(x+y+z)^{3}+4\lambda^{2}(x+y+z)(xy+yz+zx)+8\lambda^{3}xyz$ $\geq (2\lambda+1)(3\sqrt[3]{xyz})^{3}+4\lambda^{2}\cdot 3\sqrt[3]{xyz}\cdot 3\sqrt[3]{(xyz)^{2}}+8\lambda^{3}xyz$ $=\left\{27(2\lambda+1)+36\lambda^{2}+8\lambda^{3}\right\}xyz$ $=(8\lambda^{3}+36\lambda^{2}+54\lambda+27)xyz$ $=(2\lambda+3)^{3}xyz$ But $xyz\le 1$ I didn't understand your solution Unsolved for 14 years??

Bump.......

I don't think this problem is true. In particular, let $n=2$, $a=(0,0)$, $b=(1,0),c=(-1,0)$. Then, $a+b+c=0$, $|a|=0,|b|=|c|=1$. Then, we substitute into our given to get \[\frac{2}{2}\cdot\frac{1+2\lambda+1}{2}\cdot\frac{1+2\lambda+1}{2}\geq(2\lambda+3)^3\iff (\lambda+1)^2\geq(2\lambda+3)\]clearly false for $\lambda\geq 100$.