# A True Algebraic-Geometric Inequality

### Solution 1

Consider points $A(0,\ldots,0),~M(a_1,\ldots,a_n),~N(b_1,\ldots,b_n)\in\mathbb{R}^n,$ the Euclidean space of dimension $n.$

If $M=A,$ or $N=A,$ or $M=N,$ there is nothing to prove.

We consider two cases:

Case 1: $A,M,N$ are collinear, so, e.g., $b_k=\alpha a_k,$ $k=1,2,\ldots,n$, $\alpha \ne 0.$

The required inequality reduces to

$\displaystyle |2-\alpha|\sqrt{\sum_{k=1}^na^2_k}+|2\alpha-1|\ge (|\alpha|+1)\sqrt{\sum_{k=1}^na^2_k},$

which is simplified to $|2-\alpha|+|2\alpha -1|\ge |\alpha|+1.$ The latter is equivalent to

$\begin{cases} |2-\alpha|+|2\alpha-1|\ge \alpha+1,&\textit{if }\alpha\ge 0,\\ (2-\alpha)+(1-2\alpha)\ge -\alpha,&\textit{if }\alpha\lt 0. \end{cases}$

The latter resolves into $2\ge 2\alpha,$ which is obviously true for $\alpha\lt 0.$

The former, by the triangle inequality for the absolute value, is also true:

$|2-\alpha|+|2\alpha-1|\ge |(2-\alpha)+(2\alpha-1)|=|1+\alpha|=1+\alpha,$

since $\alpha\ge 0.$

Case 2: $A,M,N$ are pairwise distinct and not collinear.

Let $B$ be the reflection of $A$ in $M$ and $C$ $A's$ reflection in $N.$ $B=(2a_1,2a_2,\ldots,2a_n)$ and $C=(2b_1,2b_2,\ldots,2b_n).$ Consider $\Delta ABC$ in which $M$ is the midpoint of $AB$ and $N$ the midpoint of $AC.$ Further, $\displaystyle AB=2\sqrt{\sum_{k=1}^na^2_k}=2\cdot AM,$ $\displaystyle AC=2\sqrt{\sum_{k=1}^nb_k^2}=2\cdot AN,$ $\displaystyle BN=\sqrt{\sum_{k=1}^n(2a_k-b_k)^2},$ and $\displaystyle CM=\sqrt{\sum_{k=1}^n(2b_k-a_k)^2}.$

By the triangle inequality in triangles $ABN$ and $ACM,$ $BN+AN\gt AB=2AM$ and $AM+CM\gt AC=2AN$ which, when added, give

$BN+AN+AM+CM\gt 2AM+2AN,$

or, $BN+CM\gt AM+AN,$ exactly what is required.

### Solution 2

The four terms of the inequality are euclidean norms:

$||2A-B||+||2B-A||\ge ||A||+||B||.$

By the triangle inequality, $||2A-B||+||B||\ge 2||A||$ while $||2B-A||+||A||\ge 2||B||.$ Adding the two delivers the required inequality.

### Acknowledgment

Lorian Saceanu has kindly communicated to me this problem and Leo Giugiuc a solution via my facebook account. The problem and the solution are by Lorian Saceanu (Romania), Leonard Giugiuc (Romania), Kadir Altintas (Turkey).