Inequality involving norms in $\mathbb{R}^{n}$. Let $x\in\mathbb{R}^{n}$.
Then $x=x_{1}e_{1}+\dots+x_{n}e_{n}$.
If $||x||$ is any norm in $\mathbb{R}^{n}$, is it true that
$$
\lVert x_{j}e_{j}\rVert\le\left\lVert\sum_{k=1}^{n}x_{k}e_{k}\right\rVert=\lVert x\rVert
$$
where $1\le j\le n$?
 A: My comment seems to be engendering some confusion, so I'll spell out explicitly what I mean. Pick any real $t \neq 0$ and take the norm
$$||y_1 e_1 + y_2 (e_1 + t e_2)|| = y_1^2 + y_2^2.$$
Then we find that 
$$|| t e_2 || = 2 > || e_1 +t e_2 || = 1.$$

If you're willing to change norms, there's no reason you shouldn't also be willing to change bases. For any norm $||v||$ and any invertible linear transformation $B : \mathbb{R}^n \to \mathbb{R}^n$, considering the new norm $||Bv||$ (as a numerical function of the coefficients $x_i$ with respect to the standard basis $e_1, ... e_n$) is equivalent to working with the old norm but as a function of the coefficients with respect to the new basis $B e_1, B e_2, ... B e_n$. 
So let's work with the ordinary Euclidean norm on $\mathbb{R}^2$ but use arbitrary basis vectors $f_1, f_2$. Then the inequality $||x_1 f_1|| \le ||x_1 f_1 + x_2 f_2||$ is clearly false if, say, $x_1 = 1, x_2 = -1$ and $f_1, f_2$ are unit vectors which are close to parallel. Translating this counterexample into a changed norm instead of a changed basis gives the above. 
A: To expand on (what I think is) Qiaochu's answer, for any norm $|| \cdot ||$, and injective linear transformation $A$, we can define a new norm
$$||Ax||$$
I think what Qiaochu is suggesting is taking the standard basis $e_{1}$ and $e_{2}$ for $\mathbb{R}^2$ with the Euclidean norm. Let $A$ be the injective transformation 
$$A(e_{1}) = e_{1}$$
$$A(e_{2}) = e_{1} + 0.0001e_{2}$$
Then, considering the norm $||Ax||$, we see that the inequality does not hold for $e_{1} - e_{2}$.
