# Real inequality

I have to prove this inequality:

Let $p$ be a real number such that $p\geq 1$. For all $\epsilon > 0$ there is a constant $C_\epsilon >0$ such that for all nonnegative real numbers $a$, $b$, \begin{equation} (a+b)^p \leq (1 + \epsilon)a^p + C_\epsilon b^p. \end{equation} Can someone help me? Thank you.

• What about $p$? Is it given and real positive? – Robert Z Nov 17 '17 at 8:14
• $p \ge 1$ real. – Skills Nov 17 '17 at 9:02
• OK. I edited your question. See my answer below. – Robert Z Nov 17 '17 at 9:04

## 1 Answer

Hint. If $b=0$ then the inequality is trivial. If $b>0$ then $$(a+b)^p \leq (1 + \epsilon)a^p + C_\epsilon b^p\Leftrightarrow f(t) \leq C_\epsilon$$ where $t=a/b\geq 0$ and $f(t)=(t+1)^p -(1 + \epsilon)t^p$.

Note that $f$ is continuous in $[0,+\infty)$. What is $\lim_{t\to +\infty} f(t)$?

• The limis is $- \infty$ – Skills Nov 17 '17 at 9:33
• Yes, this means that $f$ is upper bounded in $[0,+\infty)$ by some positive constant $C_{\epsilon}$. Is it clear? – Robert Z Nov 17 '17 at 9:42
• Yes. Thank you very much – Skills Nov 17 '17 at 10:03