Projection of a point into the ellipsoid and the related inequality

Question

Is this inequality true? I tried to prove it using the Cauchy–Schwarz inequality but it is not simple for me. $$ \sum_{k=1}^{n}c_k^2 z_k^2 \bigg(\sum_{k=1}^{n}x_k^2\bigg)^2 \geq \sum_{k=1}^{n}c_k^2 x_k^2 \bigg(\sum_{k=1}^{n}x_k z_k\bigg)^2 $$ This inequality has appeared from my problem to find a projection of a point $x=(x_1, x_2, \ldots, x_n)\in\mathbb{R}^n$ into the ellipsoid $$ E=\bigg\{x\in\mathbb{R}^n: \, \sum_{k=1}^{n}c_k^2 x_k^2 \leq R^2\bigg\}. $$ I think that the projection of a point $x\in\mathbb{R}^n\setminus E$ is $y\in\mathbb{R}^n$, $y=(y_1, y_2, \ldots, y_n)$, $$ y_k=\frac{Rx_k}{\sqrt{\sum\limits_{j=1}^{n}c_j^2 x_j^2}}, \quad k=1,2,\ldots,n. $$ Am I right or not? I have proved that such $y$ belongs to the ellipsoid $E$. And I have to prove that the inner product $(y-x,z-y)\geq 0$ $\forall\,z\in E$. And my inequality has appeared from this problem when I tried to prove this.

Answer 1

Not true. Take:
$(x1, x2, z1, z2, c1, c2) = (1,0, 1/c1 , 0, c1, anything)$
then the inequality reduces to $c1 \le 1$, which can be easily violated taking $c1 = 2$ are any other sufficienty large number.