Cần cù bù thông minh vậy 
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ĐK: $x\geq \frac{-5}{2}$
BPT $\Leftrightarrow (x^3+3x^2+14x+15-2(x+2)\sqrt{2x+5}-3(x+2)\sqrt{x^2+5}-\sqrt[3]{5x^2+7}\leq 0$
$\Leftrightarrow x^3+3x^2-x-18-2(x+2)(\sqrt{2x+5}-3)-3(x+2)(\sqrt{x^2+5}-3)+3-\sqrt[3]{5x^2+7}\leq 0$
$\Leftrightarrow $ Liên hợp ~~
$\Leftrightarrow (x-2)$$[x^2+5x+9-\frac{4(x+2)}{\sqrt{2x+5}+3}-\frac{3(x+2)^2}{\sqrt{x^2+5}+3}-\frac{5(x+2)}{9+3\sqrt[3]{5x^2+7}+(\sqrt[3]{5x^2+7})^2}$$\leq 0$
Với $x\geq \frac{-5}{2}\Rightarrow \left\{ \begin{array}{l} \frac{4(x+2)}{\sqrt{2x+5}+3}< \frac{4}{3}(x+2)\\ \frac{3(x+2)^2}{\sqrt{x^2+5}+3}<\frac{3}{5}(x+2)\\\frac{5(x+2)}{..........}<\frac{5(x+2)}{9} \end{array} \right.\Rightarrow$ $f(x)$ $>\frac{18x^2+57x+127}{45}>0$
Do đó $x-2\leq 0\Leftrightarrow x\leq 2$
Kết hợp đc: $x \in [\frac{-5}{2};2]$