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I am wondering if we have the following minoration of the $\max$ function :

$$ \forall a, b, c \in \mathbf{R} ~~~~~ \max(a, b, c) \geq \dfrac{1}{3} ( a+b+c) $$

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3 Answers 3

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$$\dfrac{1}{3} ( a+b+c) \leqslant \frac{\max(a,b,c)+\max(a,b,c)+\max(a,b,c)}{3} = \max(a,b,c)$$ $$\min(a,b,c) = \frac{\min(a,b,c)+\min(a,b,c)+\min(a,b,c)}{3} \leqslant \dfrac{1}{3} ( a+b+c)$$ And so for any fixed amount of numbers. $$\min(a_1,\cdots,a_n)\leqslant \frac{1}{n}\sum_{i=1}^{n}a_i\leqslant \max(a_1,\cdots,a_n)$$

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hint

By definition

$$\max(a,b,c)\ge a$$ $$\max(a,b,c)\ge b$$ $$\max(a,b,c)\ge c$$

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Also, $\max(a,b,c) = \max_{\sum_k \lambda_k = 1, \lambda_k\ge0} (\lambda_1 a + \lambda_2 b + \lambda_3c ) \ge {1 \over 3 } a +{1 \over 3 } b +{1 \over 3 } b$.

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