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Doan Quynh
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Nguyễn Hưng
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Phạm Thiên Trang
10 tháng 2 2017 lúc 22:56

chán quá

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Phạm Thiên Trang
10 tháng 2 2017 lúc 22:57

Chẳng có bài toán nào cả .Các bạn giải hết rồi

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Nguyễn Hưng
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Uyên Hoàng
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toán khó mới hay
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alibaba nguyễn
11 tháng 11 2017 lúc 8:25

Ta có:

\(\frac{a+1}{1+b^2}=a+1-\frac{\left(a+1\right)b^2}{1+b^2}\ge a+1-\frac{\left(a+1\right)b^2}{2b}=a+1-\frac{ab+b}{2}\left(1\right)\)

Tương tụ ta có:

\(\hept{\begin{cases}\frac{\left(b+1\right)}{1+c^2}\ge b+1-\frac{bc+c}{2}\left(2\right)\\\frac{\left(c+1\right)}{1+a^2}\ge c+1-\frac{ca+a}{2}\left(3\right)\end{cases}}\)

Từ (1), (2), (3) ta có:

\(M\ge a+b+c+3-\frac{ab+bc+ca+a+b+c}{2}\)

\(=3+3-\frac{ab+bc+ca+3}{2}\)

\(\ge\frac{9}{2}-\frac{\left(a+b+c\right)^2}{6}=3\)

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Hiếu Lê
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Tran Le Khanh Linh
20 tháng 8 2020 lúc 20:17

Chắc áp dụng được Cauchy-Schwarz

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Kiệt Nguyễn
24 tháng 11 2020 lúc 20:42

Ta có: \(\sqrt[3]{\left(a+b\right).\frac{2}{3}.\frac{2}{3}}\le\frac{a+b+\frac{4}{3}}{3}=\frac{a+b}{3}+\frac{4}{9}\)

Tương tự rồi cộng các vế của BĐT lại, ta được: \(\sqrt[3]{\frac{4}{9}}P\le\frac{2\left(a+b+c\right)}{3}+\frac{4}{3}=2\Rightarrow P\le\sqrt[3]{18}\)

Đẳng thức xảy ra khi \(a=b=c=\frac{1}{3}\)

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lỡ k Kiệt Nguyễn r, bài Kiệt Nguyên sai r

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Kim Khánh Linh
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Bellion
15 tháng 5 2021 lúc 14:30

                      Bài làm :

Ta có :

\(\left(a+b\right)^2\ge4ab\)

\(\Leftrightarrow\frac{a+b}{ab}\ge\frac{4}{a+b}\)

\(\Leftrightarrow\frac{4}{a+b}\le\frac{1}{a}+\frac{1}{b}\)

\(\Leftrightarrow\frac{1}{a+b}\le\frac{1}{4}\left(\frac{1}{a}+\frac{1}{b}\right)\left(1\right)\)

Dấu "=" xảy ra khi : a=b

Chứng minh tương tự như trên ; ta có :

\(\hept{\begin{cases}\frac{1}{b+c}\text{≤}\frac{1}{4}\left(\frac{1}{b}+\frac{1}{c}\right)\left(2\right)\\\frac{1}{c+a}\text{≤}\frac{1}{4}\left(\frac{1}{c}+\frac{1}{a}\right)\left(3\right)\end{cases}}\)

Cộng vế với vế của (1) ; (2) ; (3) ; ta được :

\(A\text{≤}\frac{1}{2}\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)\text{=}\frac{3}{2}\)

Dấu "=" xảy ra khi ;

\(\hept{\begin{cases}a=b=c\\\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=3\end{cases}}\Leftrightarrow a=b=c=1\)

Vậy Max (A) = 3/2 khi a=b=c=1

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Ối giời ối giời ôi
15 tháng 5 2021 lúc 14:14

quản lí tên kiểu j z

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Ối giời ối giời ôi
15 tháng 5 2021 lúc 14:14

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Thùy Trang Hoàng
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Chu Phương Uyên
6 tháng 3 2017 lúc 21:17

Kiểm tra mà bạn vẫn có thời gian đưa câu hỏi ư! Bái phục mà thi j vậy bn?

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Pham Trong Bach
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Cao Minh Tâm
15 tháng 9 2019 lúc 2:22

Vì:  a + 1 1 + b 2 = a + 1 − b 2 ( a + 1 ) 1 + b 2 ;   1 + b 2 ≥ 2 b   n ê n   a + 1 1 + b 2 ≥ a + 1 − b 2 ( a + 1 ) 2 b = a + 1 − a b + b 2

Tương tự:  b + 1 1 + c 2 ≥ b + 1 − b c + c 2 ;   c + 1 1 + a 2 ≥ c + 1 − c a + a 2 ⇒ M ≥ a + b + c + 3 − ( a + b + c ) + ( a b + b c + c a ) 2 = 3 + 3 − ( a b + b c + c a ) 2

Chứng minh được:  3 ( a b + b c + c a ) ≤ ( a + b + c ) 2 = 9 a c ⇒ 3 − ( a b + b c + c a ) 2 ≥ 0 ⇒ M ≥ 3

Dấu “=” xảy ra khi a = b = c = 1. Giá trị nhỏ nhất của M bằng 3.

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