Assume that $p \geq 2$. Show the following equation: $x^p+y^p\leqslant (x^2+y^2)^{p/2}$
What is the right solution of this problem?
For $p=2$ it's easy.
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Blabla
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What have you tried? – Calvin Lin Dec 08 '19 at 18:32
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Check your source. Do you want $p \geq 2$? E.g. $p=1, x = 3, y = 4$ is false. – Calvin Lin Dec 08 '19 at 18:33
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Yes, I assume that $p \geq 2$ – Blabla Dec 08 '19 at 18:38
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4Then do not assume it in comments but rather include it in the main post, together with some more context (for example what have you tried). – Sil Dec 08 '19 at 18:49
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1Maybe this helps: https://math.stackexchange.com/q/4094/631742 – Maximilian Janisch Dec 08 '19 at 19:23
1 Answers
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We shall prove that $\frac{d}{dp}(x^p+y^p)^{\frac{1}{p}} \leq 0$, and this will prove the result.
Note that $\frac{d}{dp}(x^p+y^p)^{\frac{1}{p}} = \frac{(x^p+y^p)^{\frac{1}{p}}}{p^2(x^p+y^p)}(x^p \ln(x^p)+y^p \ln y^p - (x^p+y^p)\ln (x^p+y^p))$.
To prove it is negative, note that the expression inside the brackets is $f(x^p+y^p)-f(x^p)-f(y^p)$ where $f(z) = z \ln z$.
$f$ is convex everywhere so we establish the result.
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