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I have that $\Phi:\mathbb{R}\rightarrow [0,+\infty)$ is a continuous, increasing and convex function such that $$t^l\Phi(1)\leq \Phi(t)\leq t^m \Phi(1), \;\;\forall t\geq1$$ where $1<l<m$.

and $\displaystyle\lim_{t\rightarrow0}\frac{\Phi(t)}{t}=0~\text{and}~\lim_{t\rightarrow+\infty}\frac{\Phi(t)}{t}=+\infty.$

I want to prove that $$t^l\leq \Phi(t)+1,\;\;\forall t\geq1$$

Can someone help me?

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This is not true. Take $l = 2$, $m = 3$ and $\Phi(t) = \frac12 \, t^2$. It is clear that your assumptions are satisfied. However, with $t = 2$ you get $$4 = t^2 \not\le \Phi(t) + 1 = \frac12 \, 2^2 + 1 = 3.$$

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