The geometric and arithmetic means of $x\in R^n_+$ are, respectively, $$G(x)=(\prod^n_{i=1}x_i)^{1/n},\quad A(x)=\frac{1}{n}\sum^n_{i=1}x_i$$ Suppose $0\leq \alpha\leq 1$, how to prove the function $G(x)-\alpha A(x)$ is concave? I try to solve it by Hessian matrix, but it seems not work.
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The geometric mean is not that you defined... – Mostafa Ayaz Mar 24 '19 at 07:57
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Sorry, I made a mistake. – Noilyn Mar 24 '19 at 08:06
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1Seems quite easy. The geometric mean is concave. Any linear function, including $-\alpha A(x)$ is concave. A sum of concave functions is concave. Am I missing something? – Alex Shtoff Mar 24 '19 at 08:37
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@AlexShtof Yeah，you are right. Thankssss! (I'm so stupid) – Noilyn Mar 24 '19 at 08:40