<s>[INST] Inverse trigonometric function problem I recently came across a problem:

If $\mathbb{sin^{-1}x+sin^{-1}y+sin^{-1}z=\pi}$ then prove that $\mathbb{x^4+y^4+z^4+4x^2y^2z^2=2(x^2y^2+y^2z^2+x^2z^2)}$

Now, I proceeded like this:
Let, $\mathbb{sin^{-1}x=A, sin^{-1}y=B, sin^{-1}z=C}$ Then, $\mathbb{x=sinA, y=sinB, z=sinC}$ and $\mathbb{A+B+C=\pi}$
If I put the values of x, y and z in the left hand side and the right hand side of the identity, both sides should be equal. But I cannot proceed anymore. Can anyone help me solve the problem?
 [/INST] My idea is to express this based on a triangle $\triangle ABC$, where $x=\sin A$, $y = \sin B$, $z=\sin C$, and let $a,b,c$ be the lengths of $BC, AC, AB$ respectively. From Heron's law, $S:=$ the area of $\triangle ABC$ can be written as $$S=\frac{1}{4}\sqrt{(a+b+c)(a+b-c)(a-b+c)(-a+b+c)}$$
and hence, $$a^4+b^4+c^4+16S^2=2\left(a^2b^2+a^2c^2+b^2c^2\right)$$
which can be proved by plugging $S$ in the left hand side and cancelling out terms.
Next, we know that in a triangle, $\ S = \frac{1}{2}bc\sin A$ and $\sin A = \frac{a}{2R}$. Hence, the left hand side can be written as $$a^4+b^4+c^4+16S^2=a^4+b^4+c^4+4b^2c^2\sin^2A=a^4+b^4+c^4+\frac{a^2b^2c^2}{R^2}$$
which gives us
$$a^4+b^4+c^4+\frac{a^2b^2c^2}{R^2} = 2\left(a^2b^2+a^2c^2+b^2c^2\right)$$
Next, we divide both sides by $16R^4$ to get:
$$\left(\frac{a}{2R}\right)^4+\left(\frac{b}{2R}\right)^4+\left(\frac{c}{2R}\right)^4+4\cdot\left(\frac{a}{2R}\right)^2\left(\frac{b}{2R}\right)^2\left(\frac{c}{2R}\right)^2 \\= 2\left(\left(\frac{a}{2R}\right)^2\left(\frac{b}{2R}\right)^2+\left(\frac{a}{2R}\right)^2\left(\frac{c}{2R}\right)^2+\left(\frac{b}{2R}\right)^2\left(\frac{c}{2R}\right)^2\right)$$
and use $x=\sin A = \frac{a}{2R}$, $y=\sin B = \frac{b}{2R}$, $z=\sin C = \frac{c}{2R}$ into the equation to get
$$x^4+y^4+z^4+4x^2y^2z^2=2\left(x^2y^2+x^2z^2+y^2z^2\right)$$
Note: The range of $\sin^{-1}$ is $\left[-\frac{\pi}{2},\frac{\pi}{2}\right]$, so with the condition $\sin^{-1}x+\sin^{-1}y+\sin^{-1}z=\pi$, the only case where we can't use a triangle to represent the problem is when one of the $\sin^{-1}x = 0$. In this case, the left hand side and right hand side both equal $2\cdot \left(\frac{\pi}{2}\right)^4$, and equality is maintained.
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