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.
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are x,y natural numbers? – teddybear May 13 '14 at 16:25
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No @teddybear, they can be any real positive number – LapLace May 13 '14 at 16:26
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Yes. $\frac{x^n +y^n}{2} \geq \left(\frac{x+y}{2}\right)^n. $ – Raghav May 13 '14 at 16:26
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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. – Prism May 13 '14 at 16:36
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Sorry, using (midpoint) convexity of $f(x)=x^n$ shows Raghav's inequality which is sharp because of $x=y=1$. – PVAL-inactive May 13 '14 at 17:04
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For a proof of Raghav's assertion ($C = 2^{n-1}$), see my answer here. – TonyK Jul 20 '14 at 11:59
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$$(x+y)^n\leq(2\operatorname{Max}(x,y))^n\leq2^n(x^n+y^n)$$
PVAL-inactive
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1@SandeepSilwal : this is straightforward. Perhaps a line to explain the last inequality would help make people understand. – user88595 May 13 '14 at 16:42
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1@SandeepSilwal: Either $\operatorname{Max}(x, y)^{n}\leq x^{n}$ or $\operatorname{Max}(x, y)^{n}\leq y^{n}$… So in any case, $\operatorname{Max}(x, y)^{n}\leq x^{n}+y^{n}$. – Prism May 13 '14 at 17:33