If $-1 < a < 1$ then $\sqrt[4]{1-a^2} + \sqrt[4]{1-a} + \sqrt[4]{1+a} < 3$?

Question

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 :) .

Answer 1

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!

Answer 2

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