# proof of a simple(?) inequality

For a fixed value of $a<\frac{1}{4}$ prove that

$\ |x|^{2a} < c_1(a) \frac{1+|x|}{1+|x|^{1-2a}}$

holds for all $x\in\mathbb R$

Similarly show that for fixed $a<\frac{1}{2}$

$\ |x|^{2a} < c_2(a) \frac{1+x^2}{1+|x|^{2-2a}}$

holds for all $x\in\mathbb R$

"i have encountered this inequality in existence result of fully nonlinear evolutionary Navier Stokes Equations..Exactly at : Navier Stokes Equations Theory and Numerical Analysis by Roger Temam pages;277,286. Thanks for your interest indeed." ($c_1(a)$ and $c_2(a)$ are constants depending on a)

• Do you mean $x\in\mathbb R$ rather than $a\in\Re$? – Andrés E. Caicedo Nov 5 '12 at 16:41
• Andres: $\mathfrak{R}$ is a (not un)common notation for the real numbers. The OP almost certainly meant $x$ instead of $a$, though. – Cameron Buie Nov 5 '12 at 17:28
• I don't want to be rude of course .. i have used only mathematical term..i have encountered this inequality in existence result of fully nonlinear evolutionary Navier Stokes Equations..Exactly at : Navier Stokes Equations Theory and Numerical Analysis by Roger Temam pages;277,286. Thanks for your interest indeed. – noname Nov 5 '12 at 17:28
• It seems the proposer and the solver know what is $c_1(a)$, but I don't. – GEdgar Nov 5 '12 at 17:54

$$\ |x|^{2a} < c_1(a) \frac{1+|x|}{1+|x|^{1-2a}} \Leftrightarrow \ |x|^{2a} (1+|x|^{1-2a}) < c_1(a) ( 1+|x|)$$

$$\Leftrightarrow \frac{\ |x|^{2a} +|x|}{1+ |x|} < c_1(a)$$

Since $\lim_{x \to \pm \infty} \frac{\ |x|^{2a} +|x|}{1+ |x|} =1$, the inequality

$$\frac{\ |x|^{2a} +|x|}{1+ |x|} <2$$

holds for all $|x| > x_0$, for some $x_0$.

The function $\frac{\ |x|^{2a} +|x|}{1+ |x|}$ is continuous on $[-x_0,x_0]$, thus has an absolute max M on this interval.

Thus we get

$$\frac{\ |x|^{2a} +|x|}{1+ |x|} <2 \,;\, \forall |x| > x_0$$ $$\frac{\ |x|^{2a} +|x|}{1+ |x|} <M \,;\, \forall |x| \leq x_0$$

This shows that $c_1(a) =\max \{2,M \}$ works.

I think the second one is wrong as stated. Note that

$$\lim_{x \to \infty} \frac{ |x|^{2}+|x|^{4-2a}}{1+|x|^2} = \infty$$

$$\ |x|^{2a} < c_2(a) \frac{1+x^2}{1+|x|^{2-2a}}$$