If $-1 < a < 1$ then $\sqrt[4]{1-a^2} + \sqrt[4]{1-a} + \sqrt[4]{1+a} < 3$? 
Prove that if $-1<a<1$, then :
$\sqrt[4]{1-a^2} + \sqrt[4]{1-a} + \sqrt[4]{1+a} < 3 $

I do not know how to even approach this problem. Those fourth roots confuse me a lot.
Any help would be appreciated.
Also I would like to know how the left hand side of the inequality behaves if there are no given bounds for the value of $a$.
Thanks in advance :) .

Let $f:[-1,1] \to \mathbb{R}$, $f(x)=\sqrt[4]{1-x}+\sqrt[4]{1+x}+\sqrt[4]{1-x^2}$. 
$f$ is obvously continous and differentable on $(-1,1)$
Then:
$$f'(x)=\frac{\sqrt[4]{(1-x)^3}-\sqrt[4]{(1+x)^3}-2x}{4\sqrt[4]{(1-x^2)^3}}$$
Of course $4\sqrt[4]{(1-x^2)^3} \geq 0$ for all $x\in[-1,1]$.


*

*for $x>0$ :
$\sqrt[4]{(1-x)^3}<\sqrt[4]{(1+x)^3}$ and $-2x<0$, so $f'(x)<0$ and $f$ is decreasing

*for $x<0$ :
$\sqrt[4]{(1-x)^3}>\sqrt[4]{(1+x)^3}$ and $-2x>0$, so $f'(x)>0$ and $f$ is increasing

*for $x=0$ : $f'(x)=0$ and $f(x)=3$


Now we can deduce, that $$\forall{x\in [1,1]}: f(x)\leq 3$$
Bounds for $a$ are required, because the number under fourth root must be nonnegative
