Chủ đề này có 1 trả lời

[Lin Kon]

Cho $a,b,c$ dương thỏa mãn $(a+b+2c)(b+c+2a)(c+a+2b)=1$.CMR:

$A=\frac{a}{b(4c+15)(b+2c)^2}+\frac{b}{c(4a+15)(c+2a)^2}+\frac{c}{a(4b+15)(a+2b)^2}\geq \frac{1}{3}$

 

[NTA1907]

Cho $a,b,c$ dương thỏa mãn $(a+b+2c)(b+c+2a)(c+a+2b)=1$.CMR:

$A=\frac{a}{b(4c+15)(b+2c)^2}+\frac{b}{c(4a+15)(c+2a)^2}+\frac{c}{a(4b+15)(a+2b)^2}\geq \frac{1}{3}$

$A=\sum \frac{a}{b(4c+15)(b+2c)^{2}}=\sum \frac{(\frac{a}{b+2c})^{2}}{ab(4c+15)}=\frac{(\frac{a^{2}}{ab+2ac})^{2}}{12abc+15(ab+bc+ca)}\geq \frac{(\frac{(a+b+c)^{2}}{3(ab+bc+ca)})^{2}}{12abc+15(ab+bc+ca)}\geq \frac{1}{12abc+15(ab+bc+ca)}$

Mà $1=\left [ (a+c)+(b+c) \right ]\left [ (a+c)+(b+a) \right ]\left [ (a+b)+(b+c) \right ]\geq 8(a+b)(b+c)(c+a)$(theo AM-GM)

Ta có bổ đề: $(a+b)(b+c)(a+c)\geq \frac{8}{9}(a+b+c)(ab+bc+ca)$

$\Rightarrow 1\geq \frac{64}{9}(a+b+c)(ab+bc+ca)\geq \frac{64}{9}.\sqrt{3(ab+bc+ca)}.(ab+bc+ca)=\frac{64}{9}.\sqrt{3(ab+bc+ca)^{3}}$

$\Rightarrow ab+bc+ca\leq \frac{3}{16}$

Ta có: $1\geq 8(a+b)(b+c)(c+a)\geq 64abc\Rightarrow abc\leq \frac{1}{64}$

$\Rightarrow A\geq \frac{1}{12.\frac{1}{64}+15.\frac{3}{16}}=\frac{1}{3}$