ĐK;$x\geq-1$
pt $\Leftrightarrow (9x+12)\sqrt{6x+8}-(19x+26)\sqrt{x+1} \leq \sqrt{(x^{2}+2x-3)^{3}} +\sqrt{x^{2}+2x-3}$ (1)
Đặt $\sqrt{6x+8}=a;\sqrt{x+1}=b \Rightarrow 9x+12=a^{2}+3b^{2}+1;19x+26=3a^{2}+b^{2}+1$
(1) $\Leftrightarrow (a^{2}+3b^{2}+1)a-(3a^{2}+b^{2}+1)b\leq \sqrt{(x^{2}+2x-3)^{3}}+\sqrt{x^{2}+2x-3}$
$\Leftrightarrow (a-b)^{3}+(a-b)\leq \sqrt{(x^{2}+2x-3)^{3}}+\sqrt{x^{2}+2x-3}$
$\Leftrightarrow (a-b-\sqrt{x^{2}+2x-3})(a-b+(a-b)\sqrt{x^{2}+2x-3}+x^{2}+2x-3+1)$
$\Leftrightarrow a-b\leq \sqrt{x^{2}+2x-3}$ do (....)>0
$\Leftrightarrow \sqrt{6x+8} -\sqrt{x+1}\leq \sqrt{x^{2}+2x-3}$
$\Leftrightarrow 6x+8\leq (\sqrt{x^{2}+2x-3} +\sqrt{x+1})^{2}$
$\Leftrightarrow (\sqrt{x^{2}-1}-\sqrt{x+3})(3\sqrt{x+3}+\sqrt{x^{2}-1})\geq0$
$ \Leftrightarrow x^{2}-1 \geq x+3$
$ \Leftrightarrow x \leq \frac{1-\sqrt{17}}{2}or x\geq\frac{1+\sqrt{17}}{2}$
kh vs ĐK $\Rightarrow x\geq \frac{1+\sqrt{17}}{2}$