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[vipboycodon]

Chứng minh bất đẳng thức: $\dfrac{1}{a+b}+\dfrac{1}{b+c}+\dfrac{1}{a+c}+\dfrac{1}{2\sqrt[3]{abc}} \ge \dfrac{a+b+c+\sqrt[3]{abc}}{(a+b)(b+c)(a+c)}$

[HoangVienDuy]

 

Chứng minh bất đẳng thức: $\dfrac{1}{a+b}+\dfrac{1}{b+c}+\dfrac{1}{a+c}+\dfrac{1}{2\sqrt[3]{abc}} \ge \dfrac{(a+b+c+\sqrt[3]{abc})^{2}}{(a+b)(b+c)(a+c)}$

đề bạn gõ nhầm rồi 

ta phải cm $\left (\dfrac{1}{a+b}+\dfrac{1}{b+c}+\dfrac{1}{a+c}+\dfrac{1}{2\sqrt[3]{abc}}.  \right )(a+b)(b+c)(a+c) \geq (a+b+c+\sqrt[3]{abc})^{2}$

thật vậy: ta có $(a+b)(b+c)(c+a)=\sum a^{2}(b+c)+2abc$

áp dụng bđt bunyacowsky ta có: $\left ( \frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{a+c}+\frac{1}{2\sqrt[3]{abc}} \right )\left [ c^{2}(a+b)+a^{2}(b+c)+b^{2}(a+c)+2abc \right ] \geq (a+b+c+\sqrt[3]{abc})^{2}$

$\Rightarrow$ đpcm

Edited by HoangVienDuy, 18-04-2015 - 22:13.