Following from the proof given by @ClementC. one can strengthen the bounds using a similar approach as proving the Cauchy Schwarz inequality,
Consider, $f(x) = (y_1^2-y_2^2-\cdots -y_n^2)x^2-2(x_1y_1-x_2y_2-\cdots-x_ny_n)x+(x_1^2-x_2^2-\cdots -x_n^2)=(y_1x-x_1)^2-(y_2x-x_1)^2-\cdots -(y_nx-x_n)^2$. Now taking $x=\frac{x_1}{y_1}$, we get $f(\frac{x_1}{y_1})=-(y_2\frac{x_1}{y_1}-x_1)^2-\cdots -(y_n\frac{x_1}{y_1}-x_n)^2\le 0$. However the leading expression of the quadratic $f$ is positive. Therefore, $f(x)\rightarrow \infty$ as $x\rightarrow \pm\infty$. Since, $f(x)\le 0$, the equation $f(x)=0$, has one root each in the intervals $(-\infty,\frac{x_1}{y_1}]$ and $[\frac{x_1}{y_1},-\infty)$. Hence the discriminant of $f$, must be non-negative, giving $0\le (y_1^2-y_2^2-\cdots -y_n^2)(x_1^2-x_2^2-\cdots -x_n^2)\le (x_1y_1-x_2y_2-\cdots -x_ny_n)^2$.
That is, $|x_1y_1+x_2y_2+\cdots +x_ny_n-2x_1y_1|\ge \sqrt{(y_1^2-y_2^2-\cdots -y_n^2)(x_1^2-x_2^2-\cdots -x_n^2)}$
or, $x_1y_1+x_2y_2+\cdots +x_ny_n \le 2x_1y_1 - \sqrt{(y_1^2-y_2^2-\cdots -y_n^2)(x_1^2-x_2^2-\cdots -x_n^2)} \le 0$.