0
$\begingroup$

Let $a$, $b$ be two positive constants. We sure have $$ a^2+b^2\geq 2ab $$ My question: would it be possible to have an inequality like $$ a^2+b^2\geq Ca^{2+\epsilon}b^{1-\eta} $$ where $C$, $\epsilon$ and $\eta$ are some positive constant?

Thank you!

$\endgroup$
  • $\begingroup$ nvm.... I think it is impossible. Just divide both side by $a^{2+\epsilon}$ and increasing $a$ we would have a contradiction... $\endgroup$ – spatially Jul 20 '16 at 19:24
  • $\begingroup$ Why did you lose $2$? It doesn't make sense to compare with to geometric mean without $2$ $\endgroup$ – Yuriy S Jul 20 '16 at 19:41
  • $\begingroup$ @You'reInMyEye I am ok with any constant... $\endgroup$ – spatially Jul 20 '16 at 19:43
  • $\begingroup$ $a^{1+\epsilon}$ will make more sense. $\endgroup$ – i707107 Jul 20 '16 at 20:06
  • $\begingroup$ $C a^{2+\varepsilon} b^{1-\eta}$ is way bigger than $a^2+b^2$ for any $a$ big enough, so: no. $\endgroup$ – Jack D'Aurizio Jul 20 '16 at 20:08
1
$\begingroup$

The following inequality is the key toward the proof of Holder's inequality:

For $u, v\geq 0$, and $p, q>0$ with $\frac1p + \frac 1q =1$,

$$ uv\leq \frac {u^p}p+ \frac{v^q}q. $$

Substitute $$a^2 = \frac{u^p}p, \ \ \mathrm{and} \ \ b^2=\frac {v^q}q.$$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.