<s>[INST] Solving $(x^2+4x+3)\sqrt{x+2}+(x^2+9x+18)\sqrt{x+7}\geq x^3+10x^2+33x+36$ I am trying but I cannot solve the following inequation:
$(x^2+4x+3)\sqrt{x+2}+(x^2+9x+18)\sqrt{x+7}\geq x^3+10x^2+33x+36$
My attemption.
The inequation is equvalent to
$$
(x+1)(x+3)\sqrt{x+2}+(x+3)(x+6)\sqrt{x+7}\geq (x+3)^2(x+4)
$$
Since $x+2\geq 0$, dividing both sides by $x+3>0$, we have
$$
(x+1)\sqrt{x+2}+(x+6)\sqrt{x+7}\geq (x+3)(x+4)
$$
 [/INST] Let us solve
$$(x+1)\sqrt{x+2}+(x+6)\sqrt{x+7}\geq (x+3)(x+4).$$
We have
\begin{align*}
 \mathrm{LHS} - \mathrm{RHS}
&= (x + 1)(\sqrt{x + 2} - 2) + (x + 6)(\sqrt{x + 7} - 3)\\[5pt]
&\qquad + 2(x + 1) + 3(x + 6) - (x + 3)(x + 4)\\[5pt]
 &= (x + 1)(\sqrt{x + 2} - 2) + (x + 6)(\sqrt{x + 7} - 3) + (x + 4)(2 - x)\\[5pt]
 &= (x + 1)\frac{x - 2}{\sqrt{x + 2} + 2} + (x + 6)\frac{x - 2}{\sqrt{x + 7} + 3} + (x + 4)(2 - x).
\end{align*}
If $-2 \le x \le 2$, we have
\begin{align*}
 \mathrm{LHS} - \mathrm{RHS}
 &= (2 - x)\left(-\frac{x + 1}{\sqrt{x + 2} + 2} - \frac{x + 6}{\sqrt{x + 7} + 3} + x + 4\right)\\
 &= (2 - x)\left(\frac{1}{\sqrt{x + 2} + 2} - \frac{x + 2}{\sqrt{x + 2} + 2} - \frac{x + 6}{\sqrt{x + 7} + 3} + x + 4\right)\\
 &\ge (2 - x)\left(0 - \frac{x + 2}{0 + 2} - \frac{x + 6}{0 + 3} + x + 4\right)\\
 &= (2 - x)(x/6 + 1)\\
 &\ge 0.
\end{align*}
If $x > 2$, we have
\begin{align*}
 \mathrm{LHS} - \mathrm{RHS}
 &= (x - 2)\left(\frac{x + 1}{\sqrt{x + 2} + 2} + \frac{x + 6}{\sqrt{x + 7} + 3} - x - 4\right)\\
 &\le (x - 2)\left(\frac{x + 1}{0 + 2} + \frac{x + 6}{0 + 3} - x - 4\right)\\
 &= (x - 2)(-x/6 - 3/2)\\
 &< 0.
\end{align*}
Thus, the solution to the inequality is $-2 \le x \le 2$.
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