Consider the following metrics in $H_0^1(\Omega)$ with $\Omega$ a bounded domain:
$$\| u\|_\lambda=\left( \int_\Omega |\nabla u|^2+\lambda\int_\Omega u^2\right)^{\frac{1}{2}}$$ and $$\| u\|_0=\left( \int_\Omega |\nabla u|^2\right)^{\frac{1}{2}}$$ so that $$\| u\|^2_\lambda=\| u\|^2_0+\lambda |u|^2_2.$$
Satisfying $\lambda > -\lambda_1$ with $\lambda_1$ the first eigenvalue of $\Delta u+\lambda u=0$ or the optimal constant in Poincaré's inequality: $$|u|_2\leq \lambda_1^{-\frac{1}{2}}\|u\|_0$$
I am to prove there are $C_1,C_2>0$ so that $$C_1\|u\|_0\leq \| u\|_\lambda \leq C_2 \| u\|_0$$ for all $u\in H_0^1$.
The right inequality is easy to prove applying Poincaré's Inequality to $\| u\|^2_\lambda=\| u\|^2_0+\lambda |u|^2_2$; the condition $\lambda > -\lambda_1$ ensures $C_2=\sqrt{1+\frac{\lambda}{\lambda_1}}$ is well defined.
How would I go about proving the other inequality?
The case $\lambda \ge 0$ is clear.
Let $\lambda < 0$. From Poincaré, you find $$-\lambda \, |u|_2^2 \le \lambda_1 \, |u|_2^2 \le \|u\|_0^2$$ Hence, for $\alpha > 0$ $$\|u\|_0^2 \le (1+\alpha)\, \|u\|_0^2 - \alpha \, \|u\|_0^2 \le (1+\alpha) \|u\|_0^2 - \alpha \, \lambda_1 \|u\|_2^2.$$ Now, choose $\alpha$ such that $-\alpha \, \lambda_1 = (1+\alpha) \, \lambda$, i.e., $\alpha = -\lambda/(\lambda+\lambda_1)$. Then, $\alpha > 0$ since $\lambda < 0 < \lambda + \lambda_1$. This gives $$\|u\|_0^2 \le (1+\alpha) \|u\|_0^2 + (1+\alpha) \,\lambda \|u\|_2^2 = (1+\alpha) \, \|u\|_\lambda^2.$$