3 replies to this topic

[tronghieu]

Cho a,b,c là các số thực ko âm có tích bằng 1.CMR
$\dfrac{1}{\sqrt{a^2+4a+13}}+\dfrac{1}{\sqrt{b^2+4b+13}}+\dfrac{1}{\sqrt{c^2+4c+13}}\leq{\dfrac{\sqrt{2}}{2}}$

[pirate]

Ta có: $ a^2+4a+13 \geq 6a+12 = 6(a+2)$
$ \Rightarrow \sum\limits_{}^{}\dfrac{1}{\sqrt{a^2+4a+13}} \leq \sum\limits_{}^{}\dfrac{1}{\sqrt{6(a+2)}} \leq 3 \sqrt[3]{\dfrac{1}{sqrt{216(a+2)(b+2)(c+2)}}} \leq \dfrac{\sqrt{2}}{2}$

Ta có: $ a^2+4a+13 \geq 6a+12 = 6(a+2)$
$ \Rightarrow \sum\limits_{}^{}\dfrac{1}{\sqrt{a^2+4a+13}} \leq \sum\limits_{}^{}\dfrac{1}{\sqrt{6(a+2)}} \leq 3 \sqrt[3]{\dfrac{1}{sqrt{216(a+2)(b+2)(c+2)}}} \leq \dfrac{\sqrt{2}}{2}$

($(a+2)(b+2)(c+2)=abc+2(ab+bc+ca)+4(a+b+c)+8 \geq abc+2.3\sqrt[3]{(abc)^2}+4.3\sqrt[3]{abc}+8=27$)

[tronghieu]

$ \sum\limits_{}^{}\dfrac{1}{\sqrt{6(a+2)}} \leq 3 \sqrt[3]{\dfrac{1}{sqrt{216(a+2)(b+2)(c+2)}}}$

Cí này sai rồi phải là dấu $\geq\$.Đó là BDT AM-GM đó

[pirate]

Xin lỗi!!!
$\sum\limits_{}^{}\dfrac{1}{\sqrt{6(a+2)}} \leq \dfrac{1}{2\sqrt{2}} \sum\limits_{}^{}(\dfrac{1}{a+2}+\dfrac{1}{3}) \leq \dfrac{\sqrt{2}}{2}$

$ \Leftrightarrow abc+ \sum\limits_{}^{}ab -4 \geq 0$
(Bất đẳng thức cuối dễ dàng chứng minh).

Edited by pirate, 20-08-2007 - 15:47.