# a inequality similar to geometric means

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!

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

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