Prove that $4x-x^4 \leq 3, x \in \Bbb R$ How can I tackle the following inequality :  

Prove that $4x-x^4 \leq 3$, where $x$  is any real number. 

Can someone point me in the right direction?
 A: Hint:
$4x-x^4-3=-(x-1)^2((x+1)^2+2)\le0$
A: For your question, consider rearranging the inequality to $$\begin{align*}4x-x^4 &\leq 3\\4x-x^4-3 &\leq 0 \end{align*}$$
Now let's prove it. For all real $x$, consider the stationary value of the function 

(let's say I didn't read the question and do not know if the stationary value is a minimum or maximum)

$$f(x)=4x-x^4-3$$ First order condition:$$\begin{align*}f'(x)=4-4x^3&=0\\4x^3&=4\\x^3&=1\\x&=1\end{align*}$$
Second order condition:$$\begin{align*}f''(x)=-12x^2&=-12\end{align*}$$
and so we conclude the maximum value of $f$ is at $x=1$. $f(1)=4-1-3=0$. We say that $f$ can never go beyond $0$, or $f(x)\leq0$ for all real $x$. It agrees with the initial inequality and hence it is true.
A: You could also apply $\mathrm{AM}-\mathrm{GM}$
$$\frac{x^{4}+1+1+1}{4} \geq \sqrt[4]{x^{4} 1\cdot 1\cdot  1}   =|x|$$
A: We see that ,  $x^4-4x+3=x^4-2x^2+1+2x^2-4x+2=(x^2-1)^2+2(x-1)^2 \geq 0$ and so $-(x^4-4x+3)\leq 0 \implies  -x^4+4x-3 \leq 0$ and 
hence   $4x-x^4 \leq 3$
A: You can also do this by some calculus. It is clear for $x \leq 0$. $4x - x^4$ is increasing until $x = 1$, where we have $4x - x^4 = 3$ on the dot, so we have it for $x \in [0,1]$. Then it decreases afterwards, so we have it.
A: Let $y = 4x-x^4-3$. Note that $y$ has just 1 extrema: $\sqrt[3]{(\frac 4 3)}$. $y$ is not positive at this point, so as isn't at $\infty$ and$-\infty$, therefore it isn't positive at $(-\infty;~\infty)$.
