Chủ đề này có 1 trả lời

[DauKeo]

1. a,b,c thỏa mãn $\frac{1}{a+1}+\frac{35}{35+2b}\leq \frac{4c}{4c+57}$

tìm min A=a.b.c

2. a,b,c,d,A,B,C,D >0 và $\frac{a}{A}= \frac{b}{B}= \frac{c}{C}= \frac{d}{D}$

CMR: $\sqrt{aA}+\sqrt{bB}+\sqrt{cC}+\sqrt{dD}=\sqrt{(a+b+c+d)(A+B+C+D)}$

[HoangKhanh2002]

2. Cho a, b, c, d, A, B, C, D >0 và $\frac{a}{A}= \frac{b}{B}= \frac{c}{C}= \frac{d}{D}$

CMR: $\sqrt{aA}+\sqrt{bB}+\sqrt{cC}+\sqrt{dD}=\sqrt{(a+b+c+d)(A+B+C+D)}$

1) Ta có: $\frac{4c}{4c+57}\geq \frac{1}{a+1}+\frac{35}{35+2b}\geq 2\sqrt{\frac{35}{(1+b)(35+2b)}}>0$

Mà:

$\frac{1}{a+1}+\frac{35}{35+2b}\leq \frac{4c}{4c+57}\Rightarrow \frac{1}{a+1}+1-\frac{2b}{35+2b}\leq \frac{4c}{4c+57}\Leftrightarrow \frac{1}{a+1}+1-\frac{4c}{4c+57}\leq \frac{2b}{35+2b}\Leftrightarrow \frac{2b}{35+2b}\geq \frac{1}{a+1}+\frac{57}{4c+57}\geq 2\sqrt{\frac{57}{(1+a)(4c+57)}}>0$

$\frac{1}{a+1}+\frac{35}{35+2b}\leq \frac{4c}{4c+57}\Rightarrow 1-\frac{a}{a+1}+\frac{35}{35+2b}\leq \frac{4c}{4c+57}\Leftrightarrow \frac{a}{a+1}\geq 1-\frac{4c}{4c+57}+\frac{35}{35+2b}=\frac{57}{4c+57}+\frac{35}{35+2b}\geq 2\sqrt{\frac{57.35}{(4c+57)(35+2b)}}>0$

Do đó: $\frac{8abc}{(a+1)(4c+57)(2b+35)}\geq 8\frac{35.57}{(1+a)(2b+35)(4c+57)}\Rightarrow abc\geq 1995$

2)

Đặt $t=\frac{a}{A}= \frac{b}{B}= \frac{c}{C}= \frac{d}{D}\Rightarrow a=tA,b=tB,c=tC,d=tD;t=\frac{a+b+c+d}{A+B+C+D}$

Vì vây ta có: $\sqrt{aA}+\sqrt{bB}+\sqrt{cC}+\sqrt{dD}=\sqrt{A^2t}+\sqrt{B^2t}+\sqrt{C^2t}+\sqrt{D^2t}=\sqrt{t}(A+B+C+D)=(A+B+C+D)\sqrt{\frac{a+b+c+d}{A+B+C+D}}=\sqrt{(a+b+c+d)(A+B+C+D)}$

Bài viết đã được chỉnh sửa nội dung bởi HoangKhanh2002: 14-03-2017 - 23:01