Chủ đề này có 1 trả lời

[RealCielo]

1,Cho a,b,c>0, abc=1.CMR

$\sum \frac{1}{(a+1)^2}+\frac{2}{(a+1)(b+1)(x+1)}\geq 1$

2,Cho tam giac ABC.CMR:

$(3-\frac{b+c}{a})(3-\frac{c+a}{b})(3-\frac{a+b}{c})\leq 1$

3,Cho a,b,c>0/CMR:

$\frac{1}{(b+c)^2}+\frac{1}{(c+a)^2}+\frac{1}{(a+b)^2}\geq \frac{3\sqrt{3abc(a+b+c)(a+b+c)^2}}{4(ab+bc+ca)^2}$

4,Cho x,y,z>0.CMR:

$\sum \frac{x^3}{x^3+(x+y)^3} \geq \frac{1}{3}$

[phamngochung9a]

1,Cho a,b,c>0, abc=1.CMR

$\sum \frac{1}{(a+1)^2}+\frac{2}{(a+1)(b+1)(x+1)}\geq 1$

2,Cho tam giac ABC.CMR:

$(3-\frac{b+c}{a})(3-\frac{c+a}{b})(3-\frac{a+b}{c})\leq 1$

3,Cho a,b,c>0/CMR:

$\frac{1}{(b+c)^2}+\frac{1}{(c+a)^2}+\frac{1}{(a+b)^2}\geq \frac{3\sqrt{3abc(a+b+c)(a+b+c)^2}}{4(ab+bc+ca)^2}$

4,Cho x,y,z>0.CMR:

$\sum \frac{x^3}{x^3+(x+y)^3} \geq \frac{1}{3}$

1) 

Theo nguyên lí Đi-dép-lê, giả sử $(a-1)(b-1)\geq 0 \Leftrightarrow ab+1\geq a+b$

$VT\geq \frac{1}{ab+1}+\frac{1}{(c+1)^{2}}+\frac{2}{(ab+a+b+1)(c+1)}\\=\frac{c}{c+1}+\frac{1}{(c+1)^{2}}+\frac{2}{2(ab+1)(c+1)}\\=\frac{c^{2}+c+1}{(c+1)^{2}}+\frac{c}{(c+1)^{2}}=1$

Bài viết đã được chỉnh sửa nội dung bởi phamngochung9a: 12-03-2016 - 20:32