Chủ đề này có 2 trả lời

[sharker]

$\frac{a}{b^3+ab}+\frac{b}{c^3+bc}+\frac{c}{a^3+ac}\geq \frac{3}{2}$

với a,b,c>0.    a+b+c=3

[trieuduc0101]

Ta có: S= $\frac{a}{b^{3}+ab}+\frac{b}{c^{3}+bc}+\frac{c}{a^{3}+ac}=\frac{1}{b}-\frac{b}{a+b^{2}}+\frac{1}{c}-\frac{c}{b+c^{2}}+\frac{1}{a}-\frac{a}{c+a^{2}}$

$\Rightarrow S\geq \frac{1}{a}+\frac{1}{b}+\frac{1}{c}-(\frac{b}{2\sqrt{ab^{2}}}+\frac{c}{2\sqrt{bc^{2}}}+\frac{a}{2\sqrt{ca^{2}}})$

$=\frac{1}{a}+\frac{1}{b}+\frac{1}{c}-\frac{1}{2}(\frac{1}{\sqrt{a}}+\frac{1}{\sqrt{b}}+\frac{1}{\sqrt{c}})$

Lại có: $\frac{1}{a}+\frac{1}{b}+\frac{1}{c}+3=(\frac{1}{a}+1)+(\frac{1}{b}+1)+(\frac{1}{c}+1)\geq 2(\frac{1}{\sqrt{a}}+\frac{1}{\sqrt{b}}+\frac{1}{\sqrt{c}})$

$\Rightarrow S\geq 2(\frac{1}{\sqrt{a}}+\frac{1}{\sqrt{b}}+\frac{1}{\sqrt{c}})-3-\frac{1}{2}(\frac{1}{\sqrt{a}}+\frac{1}{\sqrt{b}}+\frac{1}{\sqrt{c}})=\frac{3}{2}(\frac{1}{\sqrt{a}}+\frac{1}{\sqrt{b}}+\frac{1}{\sqrt{c}})-3$

$\geq \frac{3}{2}\frac{3}{\sqrt{\frac{a+b+c}{3}}}-3=\frac{3}{2}$

[sharker]

dấu = thứ 2 bạn sai rồi nhé... @@. tử phải là b ^2