$x^2$=$cos^2(\frac{a-b}{2})$=$\frac{cos(a-b)+1}{2}$
$\bmatrix{sina&cosa&1\\ sinb&cosb&1\\sinc&cosc&1}$$\bmatrix{sina&sinb&sinc\\ cosa&cosb&cosc\\1&1&1}$=$\bmatrix{2&2x^2&2z^2\\2x^2&2&2y^2\\2z^2&2y^2&2}$
$det\bmatrix{sina&cosa&1\\ sinb&cosb&1\\sinc&cosc&1}$$det\bmatrix{sina&sinb&sinc\\ cosa&cosb&cosc\\1&1&1}$=$det\bmatrix{2&2x^2&2z^2\\2x^2&2&2y^2\\2z^2&2y^2&2}$
จาก$det\bmatrix{sina&cosa&1\\ sinb&cosb&1\\sinc&cosc&1}$=$det\bmatrix{sina&sinb&sinc\\ cosa&cosb&cosc\\1&1&1}$
ดังนั้น $det\bmatrix{2&2x^2&2z^2\\2x^2&2&2y^2\\2z^2&2y^2&2}\ge0$
$det\bmatrix{1&x^2&z^2\\x^2&1&y^2\\z^2&y^2&1}\ge0$
$1+2x^2y^2z^2-x^4-y^4-z^4\ge0$
$\therefore1+2x^2y^2z^2\ge x^4+y^4+z^4$
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