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!

  • $\begingroup$ 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

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


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