<s>[INST] What is wrong with this solution of find the least value of $ \sec^6 x +\csc^6 x + \sec^6 x\csc^6 x$ What is wrong with this solution of find the least value of $ \sec^6 x +\csc^6 x + \sec^6 x\csc^6 x$
They all are positive terms so arithmetic mean is greater than equal to geometric mean.
$$ \sec^6 x +\csc^6 x + \sec^6 x\csc^6 x\geq  3(  \sec^6 x \csc^6 x \sec^6 x\csc^6 x)^\frac{1}{3}   $$
$$ \sec^6 x +\csc^6 x + \sec^6 x\csc^6 x \geq 3(  \sec x \csc)^4 $$
$$
\sec^6 x +\csc^6 x + \sec^6 x\csc^6 x\geq
 \frac{3 * 2^4}{\sin ^4 2x} $$
Clearly least value is 48, but something is wrong here, as the answer is 80, if I use other methods.
 [/INST] Alternatively:
$$\sec^6 x +\csc^6 x + \sec^6 x\csc^6 x=\frac{\sin^6x+\cos^6x+1}{\sin^6 x\cos^6x}=\\
\frac{2^6[(\sin^2x+\cos^2x)^3-3\sin^2x\cos^2x(\sin^2x+\cos^2x)+1]}{\sin^62x}=\\
\frac{64(2-\frac34\sin^22x)}{(\sin^22x)^3}=\frac{128-48(1-\cos^22x)}{(\sin^22x)^3}=\\
\frac{80+48\cos^22x}{(\sin^22x)^3}\ge 80.$$
equality occurs for $\sin2x=\pm1$.
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