<s>[INST] $2x^4 \le\sin^4x+\cos^6x -1 $ inequality Solve the inequality:
$2x^4 \le \sin^4x+\cos^6x -1 $
What I did:
Using trig. identities I simplified the RHS to: $3\sin^2(x)\cos^2(x) + (\sin^2(x))\cos^2(x)\sin^2(x)$ but still I don't know where to go from here. I realize that both sides are positive (at least if I move the 1 to the LHS). Wolfram says the solution is $x=0$. But I can't figure how to get there.
Thanks
 [/INST] Hint: 
Maximum value of $\sin^4(x) + \cos^6(x)$ is $1$
$$
\begin{cases}
f(x)=\sin^4(x) + \cos^6(x) \\
f(x)=(1-\cos^2(x))^2 + cos^6(x) \\
f(x)=1-2\cdot cos^2(x)+cos^4(x)+cos^6(x) \\
f(x)=cos^4(x)+cos^6(x) - 2\cdot cos^2(x) + 1\\
\text{As all the terms are in the same phase}\\
\max(f(x)) = max(cos^4(x)+cos^6(x) - 2\cdot cos^2(x)) + 1\\
\max(f(x)) =\max(\cos^4(x)) + \max(\cos^6(x)) -2\cdot \max(\cos^2(x)) +1\\
\max(f(x)) =\max(\cos(x))^4 + \max(\cos(x))^6 -2\cdot \max(\cos(x))^2 +1\\
\text{As } \max(\cos(x)) = 1 \\
\max(f(x)) = 1 + 1 - 2 + 1= 1
\end{cases}
$$
So, Maximum Value of $sin^4(x) + cos^6(x) - 1$ is $0$
if $x$ is real, minimum value of $x^4$ is $0$
Now its easy to solve the inequality 

Its always good to visualize,
Here is the plot of $sin^4(x) + cos^6(x) - 1$

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