Coffin Problem No.1

$\begin{aligned} x\left(8\sqrt{1-x}+\sqrt{1+x}\right) & \le 11\sqrt{1+x}-16\sqrt{1-x} \\ (8x+16)\sqrt{1-x} & \le (11-x)\sqrt{1+x} & \small \color{#3D99F6} \text{Squaring both sides} \\ (64x^2+256x+256)(1-x) & \le (x^2 - 22x + 121)(1+x) \\ -64x^3-192x^2 + 256 & \le x^3 -21x^2 +99x + 121 \\ 65x^3+171x^2+99x-135 & \ge 0 \\ (5x-3)\left(13x^2 + 42x+45\right) & \ge 0 \\ \implies x & \ge \frac 35 = \boxed{0.6} & \small \color{#3D99F6} 13x^2 + 42x+45\text{ has no real root.} \end{aligned}$

Let $a=\sqrt{1-x}, b = \sqrt{1+x}.$ Then $x=b^2-1, a^2+b^2 = 2,$ and the equation becomes $\begin{aligned} (b^2-1)(8a+b) &\le 11b-16a \\ b^2(8a+b) &\le 12b-8a \\ 8a(b^2+1) &\le 12b-b^3 \\ 64a^2(b^2+1)^2 &\le b^2(12-b^2)^2 \\ 64(2-b^2)(b^2+1)^2 &\le b^2(12-b^2)^2 \\ &\vdots \\ 0 &\le 65b^6-24b^4-48b^2-128 \\ 0 &\le (5b^2-8)(13b^4+16b^2+16). \end{aligned}$ The second term is always positive, so the inequality is equivalent to $5b^2-8 \ge 0,$ or $5(x+1)-8 \ge 0,$ or $x \ge 3/5.$ Hence the minimum value for $x$ is $\fbox{0.6}.$

Using the substitution

$\displaystyle y = \frac{\sqrt{1-x}}{\sqrt{1+x}}\quad \mathrm{and}\quad\displaystyle x = \frac{1-y^2}{1+y^2}$

The inequality reduces to

$(1-y^2)(8y+1)\le (11-16y)(1+y^2)$

expanding and rearranging the terms we get

$(2y-1)(2y^2-2y+5)\le 0$

and because $2y^2-2y+5=0$ has no real solutions, we know that $\displaystyle y \le \frac{1}{2}$

and finally, solving

$\displaystyle \frac{\sqrt{1-x}}{\sqrt{1+x}} \le \frac{1}{2}$

we have $x\ge \boxed{\displaystyle \frac{3}{5}}$

I found it convenient to plot it and we can see the solution is at $x=0.6$ .