Find the domain of $x$ in $4\sqrt{x+1}+2\sqrt{2x+3}\leq(x-1)(x^2-2)$

Question

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 $x\ge 3$.

Answer 1

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 $x\ge 3$.

Answer 2

After more than two years I managed to find a more simple and creative solution of this homework-looking inequality. :-)

Define the function from the left hand side of the inequality by $f(x)$ and the function from the right hand side of the inequality by $g(x)$. Draw the graphs of these functions at a segment containing the proposed key points $-1$ and $3$. Since $\sqrt{x+1}$ exists, admissible $x$ should be not less than $-1$, so as the left endpoint of the segment we pick $-1$. As its right endpoint we pick $3+1=4$.

At the picture the graph of the function $f(x)$ is drawn in red and the graph of the function $g(x)$ is drawn in blue. Note the equality of the values of $f$ and $g$ at the key points $x=-1$ and $x=3$. The picture suggests to take a linear function $h(x)$ which graph is drawn in green, separating the functions $f$ and $g$. That is $f(x)\ge h(x)\ge g(x)$ for $-1\le x\le 3$ and $f(x)\le h(x)\le g(x)$ for $x\ge 3$. The values $f(-1)=g(-1)=h(-1)=2$ and $f(3)=g(3)=h(3)=14$ uniquely determine the linear function $h(x)=3x+5$.

It remains to check that the function $h$ satisfies the required inequalities. For this purpose we use the following equivalent transformations.

Case 1. $4\sqrt{x+1}+2\sqrt{2x+3}\le 3x+5$

$16(x+1)+4(2x+3)+16\sqrt{(x+1)(2x+3)}\le 9x^2+30x+25$

$16\sqrt{(x+1)(2x+3)}\le 9x^2+6x-3$

If $9x^2+6x-3<0$ that is when $-1<x<1/3$, the inequality is false, in the opposite cases we can square both its sides.

$256(x+1)(2x+3)\le (9x^2+6x-3)^2$

$0\le 81x^4+108x^3-530x^2-1316x-759$

The right hand side looks frustrating, but recall that we have the equality when $x=-1$ and $x=3$, so these values are its roots. Dividing we obtain

$0\le (x+1)(x-3)(81x^2+270x+253)=(x+1)(x-3)((9x+15)^2+28)$.

Thus (for $x\ge -1$) the inequality

$16(x+1)+4(2x+3)+16\sqrt{(x+1)(2x+3)} \le (x-1)^2(x^2-2)^2$

holds iff $x=-1$ or $x\ge 3$.

Case 2. $3x+5\le (x-1)(x^2-2)$

$0\le x^3-x^2-5x-3$

Again dividing by $(x+1)(x-3)$ we obtain

$0\le (x+1)^2(x-3)$.

Thus (for $x\ge -1$) the inequality

$3x+5\le (x-1)(x^2-2)$

holds iff $x\ge 3$.