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$(\sum_{i=1}^n x_i)(\sum_{i=1}^n\frac{1}{x_i})\leq25+6(\sqrt{\frac{a}{b}}+\sqrt{\frac{b}{a}})^2$

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Dark Magician 2k2

Dark Magician 2k2

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Bài 1. Cho tam giác $ABC$ với $3$ cạnh $a,b,c$. Kí hiệu 

$\left\{\begin{matrix} f(a,b,c)=\frac{1}{4}(\frac{a^2}{(p-a)^2}+\frac{b^2}{(p-b)^2}+\frac{c^2}{(p-c)^2})\\ g(a,b,c)=\frac{p-a}{\sqrt{(p-b)(p-c)}}+\frac{p-b}{\sqrt{(p-b)(p-a)}}+\frac{p-c}{\sqrt{(p-a)(p-b)}} \end{matrix}\right.$

Chứng minh rằng 

$f(a,b,c)\geq max\left\{3;g(a,b,c)\right\}$

Bài 2. Cho $0<a<b$ và $x_i=[a;b],\forall i=\overline{1,5}$

Chứng minh rằng 

$(\sum_{i=1}^n x_i)(\sum_{i=1}^n\frac{1}{x_i})\leq25+6(\sqrt{\frac{a}{b}}+\sqrt{\frac{b}{a}})^2$

Hãy tổng quát bài toán cho $n$ biến.


Edited by Dark Magician 2k2, 30-12-2016 - 19:25.





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