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Let $x,y > 0$ be two real variables and let $p > 1$. Show that:

$$x^p + px^{p-1}y \; \le \; [x + y ]^p \; \le \; x^p + p[x + y]^{p-1}y$$ and $$ [x + y]^p \le 2^{p-1}(x^p + y^p$$

I've tried computing the second derivatives in an attempt to make some convexity argument. I've also played around with Young's, Holder's, and Minkowski's Inequalities, but I've been getting nowhere. Any ideas?

Thank you for your input!

  • 0
    with $[x+y]^p$ do you mean $(x+y)^p$ or something else? – 2017-02-19
  • 0
    If yes I think you can solve it using the Binomial theorem. – 2017-02-19

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Hint: try Bernoulli's inequality (generalised version), e.g. because $p>1$ then $$(1+x)^p\geq 1+px$$ thus (because $x \ne0$, because $x>0$ is given) $$(x+y)^p=x^p\left(1+\frac{y}{x}\right)^p\geq x^p\left(1+p\frac{y}{x}\right)=x^p+px^{p-1}y$$