<s>[INST] Find the domain of $x$ in $4\sqrt{x+1}+2\sqrt{2x+3}\leq(x-1)(x^2-2)$ 
Solve this equation for $x$: $4\sqrt{x+1}+2\sqrt{2x+3}\leq(x-1)(x^2-2)$

I have no idea to solve that, but I know solutions are $x=-1$ or $x\ge 3$.
 [/INST] It seems the following.
We can straightforwardly verify your answer with Mathcad as follows:
Since $\sqrt{x+1}$ exists, $x\ge -1$. Since the left-hand side of the inequality is non-negative, its right-hand side is non-negative too, that is $x\ge\sqrt{2}$ or $-1\le x\le 1$. Squaring both sides of the inequality, we obtain:
$$16(x+1)+4(2x+3)+16\sqrt{(x+1)(2x+3)} \leq(x-1)^2(x^2-2)^2$$ 
$$16\sqrt{(x+1)(2x+3)}\le x^6-2x^5-3x^4+8x^3-32x-24$$
Mathcad shows that the right-hand side has two real roots: $x_1=-1$ and $x_2\simeq 2.7472$. 

So $x\ge x_2.$
Squaring both sides of the inequality again, we obtain:
$$0 \le x^{12}-4x^{11}-2x^{10}+28x^9-23x^8-112x^7+144x^6+288x^5-368x^4-384x^3+512x^2+256x-192$$
$$0 \le (x-3)(x+1)g(x)$$
where $$g(x)=x^{10}-2x^9-3x^8+16x^7-64x^5+16x^4+128x^3-64x^2-128x+64.$$
Mathcad shows that the function $g$ has two real roots $x’_1\simeq 0.4917$ and $x_2’\simeq 1.9309$, both are less than $x_2$. 

Therefore the value of $(x-3)(x+1)g(x)$ is non-positive, if $x_2\le x\le 3$ and non-negative, if $x\ge 3$. 
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