How to prove that $(a+b)^{1/n} \le a^{1/n}+b^{1/n}$ by the convexity of $(a+b)^{1/n}$
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Welcome to MSE. Please how-to-ask-a-good-question. To receive attention, you need to show effort (what you have tried, looked up, etc) – EditPiAf Dec 04 '20 at 23:54
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Welcome to MSE: as a hint take $f(x)=\sqrt[n]{x+1}$ and prove for concavinty of $f(x)$ then use $$\sqrt[n]{x+1}\leq \sqrt[n]{x}+\sqrt[n]{1}$$ now it suffice to put $x=\frac ab$ and multiply both sides by $\sqrt[n]{b}$
extra hint: if you accept $f(x)=\sqrt[n]{x+1}$ is concave function,so $$\sqrt[n]{x+1}\leq \sqrt[n]{x}+\sqrt[n]{1}\\ \text{ put } x=\frac ab \to \\\sqrt[n]{\frac ab+1}\leq \sqrt[n]{\frac ab}+\sqrt[n]{1}\\\text{ multiply by} \sqrt[n]{b} \text{both sides} $$
Khosrotash
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sorry but I tried now to prove that f is convex and that $\sqrt[n]{x+1}\leq \sqrt[n]{x}+\sqrt[n]{1}$ but with no conclusion ... extra hint pls – sara Nov 29 '20 at 14:55
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@user854662 How did you try to prove $f$ is convex? What tools/definitions do you use? – user854214 Nov 29 '20 at 15:27
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oh sorry, this is a concave function. I write without care . I'll correct it . but the solution is true for a concave function – Khosrotash Nov 29 '20 at 15:32
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https://en.wikipedia.org/wiki/Concave_function read this about concave function – Khosrotash Nov 29 '20 at 15:33
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as a simple example of concave function see $$\sqrt{a+b}\leq \sqrt{a}+\sqrt{b}$$ – Khosrotash Nov 29 '20 at 15:34