better lower bound than $a^2t^2+b^2 \geq \sqrt{a^2+b^2}t?$

Question

If $t\geq 1,|a|,|b|\geq 1;a,b\in \mathbb{Z}$, is there a better lower bound for $a^2t^2+b^2$ than $$a^2t^2+b^2 \geq \sqrt{a^2+b^2}t.$$ "Better" in the sense, can I have a higher power of $t$ than 1? It can be even be $t^{1+\epsilon}.$ However, I need both $a$ and $b$ present in it,existing in some nice homogeneous form, separated from $t$, as above, like powers of $a^2+b^2.$ Any references also would be equally helpful. P.S: I am trying to estimate $$\displaystyle\sum_{M\geq 2} \displaystyle\sum_{a^2+b^2 = M} \left(a^2t^2+b^2\right)^{-\frac{1}{2}\sigma}$$ where $\sigma \in \mathbb{R}$ is positive and large.