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.

In your way you proved that the minimal value is greater than $48$. 
It's true, but the equality does not occur, which says that $48$ is not a minimal value.
The right solution can be the following, for example.
Let $\sin^2x\cos^2x=t.$
Thus, by AM-GM
$$t\leq\left(\frac{\sin^2x+\cos^2x}{2}\right)^2=\frac{1}{4}.$$
The equality occurs for $x=45^{\circ},$ which gives a value $80$.
We'll prove that it's a minimal value.
Indeed, we need to prove that
$$\frac{\sin^4x-\sin^2x\cos^2x+\cos^4x}{\sin^6x\cos^6x}+\frac{1}{\sin^6x\cos^6x}\geq80$$ or
$$\frac{1-3\sin^2x\cos^2x}{\sin^6x\cos^6x}+\frac{1}{\sin^6x\cos^6x}\geq80$$ or
$$\frac{2}{t^3}-\frac{3}{t^2}\geq80$$ or
$$80t^3+3t-2\leq0$$ or
$$80t^3-20t^2+20t^2-5t+8t-2\leq0$$ or
$$(4t-1)(20t^2+5t+2)\leq0,$$ which is obvious.
