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Solve the inequation: $5+3\sqrt{1-x^2}\geq x+4\sqrt{1-x}+3\sqrt{1+x}$

I tried to substitute $x=\cos t$ but don't get any result. Who can help me?

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First of all, we need to have $1-x^2\ge 0$, $1-x\ge 0$ and $1+x\ge 0$, i.e. $-1\le x\le 1$.

Now since the both sides of $$5-x+3\sqrt{1-x^2}\ge 4\sqrt{1-x}+3\sqrt{1+x}\tag 1$$ are non-negative, this is equivalent to $$\left(5-x+3\sqrt{1-x^2}\right)^2\ge \left(4\sqrt{1-x}+3\sqrt{1+x}\right)^2,$$ i.e. $$6(1-x)\sqrt{1-x^2}\ge 8x^2+3x-9\tag 2$$ Here, the solutions of $8x^2+3x-9=0$ are $x=\frac{-3\pm 3\sqrt{33}}{16}$ where $$\frac{-3-3\sqrt{33}}{16}\lt -1\lt\frac{-3+3\sqrt{33}}{16}\lt 1.$$ So, for $-1\le x\le\frac{-3+3\sqrt{33}}{16}$, since the LHS of $(2)$ is non-negative and the RHS of $(2)$ is non-positive, the given inequality $(1)$ holds.

For $\frac{-3+3\sqrt{33}}{16}\lt x\le 1$, since the both sides of $(2)$ are non-negative, $(2)$ is equivalent to $$\left(6(1-x)\sqrt{1-x^2}\right)^2\ge \left(8x^2+3x-9\right)^2,$$ i.e. $$(4x^2-3)(25x^2-6x-15)\le 0,$$ i.e. $$-\frac{\sqrt 3}{2}\le x\le \frac{3-8\sqrt 6}{25}\ \ \ \text{or}\ \ \ \frac{\sqrt 3}{2}\le x\le \frac{3+8\sqrt 6}{25}.$$ Hence, in this case, we have $\frac{-3+3\sqrt{33}}{16}\le x\le \frac{3+8\sqrt 6}{25}$.

Hence, the answer is $$-1\le x\le\frac{3+8\sqrt 6}{25}.$$

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