I have no clue on this one so I hope you can help me out on this one.
Let $x,y \ge 0 $ and $0 \lt n \le 1$. Then $$ (x+y)^n \le x^n +y^n$$
Prove that $ (x+y)^n \le x^n +y^n$ with $x,y \ge 0 $ and $0 \lt n \le 1$
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real-analysis
analysis
inequality
substitution
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1Divide by $y^n$ then remember (or prove) [Bernoulli's inequality](https://en.wikipedia.org/wiki/Bernoulli's_inequality). – 2017-01-27
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0Write the binomial expansion of the LHS, if you know how to for $n\in\Bbb R_{\ge 0}$, and compare both sides. – 2017-01-27
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0$n$ doesn't look to be integer. – 2017-01-27
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0@DietrichBurde This case is not compatible with "no clue on this one". – 2017-01-27
1 Answers
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Since our inequality is symmetric, we can assume that $y\geq x$.
Also if $x=0$ so the inequality is obviously true.
Let $x>0$ and $y=tx$, where $t\geq1$.
Hence, we need to prove that $f(t)\geq0$, where $f(t)=t^n+1-(t+1)^n$.
But, $f'(t)=n\left(\frac{1}{t^{1-n}}-\frac{1}{(1+t)^{1-n}}\right)\geq0$.
Thus, $f(t)\geq f(1)=0$ and we are done!
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0Nice Arquady, can we prove it without derivative, If yes then plz explain here, Thanks – 2017-01-28
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1@juantheron See my proof here: http://math.stackexchange.com/questions/1990936 I think we can understand convexity and Karamata without derivative. – 2017-01-28