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 :) .

It's wrong of course. Try $a=0$.
$\sqrt[4]{1-a^2} + \sqrt[4]{1-a} + \sqrt[4]{1+a}\leq3$ is true.
Indeed, by P-M $(x+y+z)^4\leq27(x^4+y^4+z^4)$. 
This inequality follows also from Holder and more:
$$(1+1+1)^3(x^4+y^4+z^4)\geq(x+y+z)^4$$
Hence, $$\left(\sqrt[4]{1-a^2} + \sqrt[4]{1-a} + \sqrt[4]{1+a}\right)^4\leq27(1-a^2+1-a+1+a)=27(3-a^2)\leq81$$
and we are done!
