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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 :) .

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    $\begingroup$ HINT: This inequality is the same as $$(\sqrt[4]{1+a}+1)(\sqrt[4]{1-a}+1)<4$$ Moreover, $-1 \le a \le 1$ is a necessary condition to ensure that those roots are real (hence well defined). $\endgroup$ – Crostul Jan 19 '17 at 7:13
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    $\begingroup$ I think there should be weak inequality. For $a=0$ the left side is equal 3. $\endgroup$ – Jaroslaw Matlak Jan 19 '17 at 7:36
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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!

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  • $\begingroup$ That was very helpful. +1 . But, can you tell me what do you mean by "P-M" ? I know a little bit about holder as perhaps the generalization of C-S, but I can not really understand what by "P-M" ??? $\endgroup$ – user399078 Jan 19 '17 at 13:57
  • $\begingroup$ @Nirbhay It's the Power Means inequality. See here: artofproblemsolving.com/wiki/index.php/Power_Mean_Inequality $\endgroup$ – Michael Rozenberg Jan 19 '17 at 14:09
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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

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