Let $n$ be a natural number. Is it possible to write $$(x+y)^n \leq C(x^n + y^n)$$ for some constant $C$??

It is obvious for $n=2$ (using Young's inequality) but not obvious to me for other $n$.

Let $x$ and $y$ be positive reals.

  • $\begingroup$ are x,y natural numbers? $\endgroup$
    – teddybear
    May 13, 2014 at 16:25
  • $\begingroup$ No @teddybear, they can be any real positive number $\endgroup$
    – LapLace
    May 13, 2014 at 16:26
  • $\begingroup$ Yes. $\frac{x^n +y^n}{2} \geq \left(\frac{x+y}{2}\right)^n. $ $\endgroup$
    – Raghav
    May 13, 2014 at 16:26
  • $\begingroup$ It would be interesting to calculate the smallest constant $C$ such that $(x+y)^n\leq C(x^n+y^n)$ holds for all positive real numbers $x$ and $y$. PVAL's answer shows $C=2^{n}$ works but perhaps there is a smaller value that works. $\endgroup$
    – Prism
    May 13, 2014 at 16:36
  • $\begingroup$ Sorry, using (midpoint) convexity of $f(x)=x^n$ shows Raghav's inequality which is sharp because of $x=y=1$. $\endgroup$ May 13, 2014 at 17:04

1 Answer 1



  • $\begingroup$ What is the proof for this please? $\endgroup$ May 13, 2014 at 16:34
  • 1
    $\begingroup$ @SandeepSilwal : this is straightforward. Perhaps a line to explain the last inequality would help make people understand. $\endgroup$
    – user88595
    May 13, 2014 at 16:42
  • 1
    $\begingroup$ @SandeepSilwal: Either $\operatorname{Max}(x, y)^{n}\leq x^{n}$ or $\operatorname{Max}(x, y)^{n}\leq y^{n}$… So in any case, $\operatorname{Max}(x, y)^{n}\leq x^{n}+y^{n}$. $\endgroup$
    – Prism
    May 13, 2014 at 17:33

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