\(\frac{1}{3a+2b+c}\le\frac{1}{36}\left(\frac{3}{a}+\frac{2}{b}+\frac{1}{c}\right)\) )cái này bn tự cm nha bằng hệ quả của bunhia
tương tự :\(\frac{1}{3b+2c+a}\le\frac{1}{36}\left(\frac{3}{b}+\frac{2}{c}+\frac{1}{a}\right)\)

\(\frac{1}{3c+2a+b}\le\frac{1}{36}\left(\frac{3}{c}+\frac{2}{a}+\frac{1}{b}\right)\)

Công tất cả các vế vs nhau:\(\frac{1}{3a+2b+c}+\frac{1}{3b+2c+a}+\frac{1}{3c+2a+b}\le\frac{1}{36}\left(\frac{6}{a}+\frac{6}{b}+\frac{6}{c}\right)\)=1/36 x96=8/3

à còn phần mik dùng bunhia sao ra dc thế nè :\(\frac{1}{3a+2b+c}=\frac{1}{a+a+a+b+b+c}\)

\(=\frac{1}{36}\left(\frac{36}{a+a+a+b+b+c}\right)\le\frac{1}{36}\left(\frac{1}{a}+\frac{1}{a}+\frac{1}{a}+\frac{1}{b}+\frac{1}{b}+\frac{1}{c}\right)\)\(=\frac{1}{36}\left(\frac{3}{a}+\frac{2}{b}+\frac{1}{c}\right)\)

tích cho tao phát thì t làm , 

TÍCH CÁI J

Theo giả thiết: \(\frac{2}{b}=\frac{1}{a}+\frac{1}{c}\ge\frac{2}{\sqrt{ac}}\Leftrightarrow b^2\le ac\Leftrightarrow\frac{ac}{b^2}\ge1\)

Ta có: \(\frac{1}{a}+\frac{1}{c}=\frac{2}{b}\Leftrightarrow b\left(a+c\right)=2ac\Leftrightarrow2ac-bc=ab\Leftrightarrow2a-b=\frac{ab}{c}\)\(\Rightarrow\frac{a+b}{2a-b}=\frac{a+b}{\frac{ab}{c}}=\frac{ac+bc}{ab}=\frac{c}{b}+\frac{c}{a}\)(1)

Tương tự: \(\frac{b+c}{2c-b}=\frac{a}{c}+\frac{a}{b}\)(2)

Cộng từng vế hai đẳng thức (1), (2) và áp dụng Cô - si, ta được: \(\frac{a+b}{2a-b}+\frac{b+c}{2c-b}\ge\frac{c}{b}+\frac{c}{a}+\frac{a}{c}+\frac{a}{b}\ge4\sqrt[4]{\frac{ca}{b^2}}\ge4\)

Đẳng thức xảy ra khi a = b = c

Ta có: 

\(\frac{1}{a}+\frac{1}{c}=\frac{2}{b}\Leftrightarrow b=\frac{2}{\frac{1}{a}+\frac{1}{c}}=\frac{2ac}{a+c}\)

Thế \(b=\frac{2ac}{a+c}\) vào M, ta được:

 \(M=\frac{a+b}{2a-b}+\frac{c+b}{2c-b}=\frac{a+\frac{2ac}{a+c}}{2a-\frac{2ac}{a+c}}+\frac{c+\frac{2ac}{a+c}}{2c-\frac{2ac}{a+c}}=\frac{1+\frac{2c}{a+c}}{2-\frac{2c}{a+c}}+\frac{1+\frac{2a}{a+c}}{2-\frac{2a}{a+c}}\)

\(M=\frac{\left(a+c\right)+2c}{2\left(a+c\right)-2c}+\frac{\left(a+c\right)+2a}{2\left(a+c\right)-2a}=\frac{a+3c}{2a}+\frac{3a+c}{2c}\)

\(M+2=\frac{a+3c}{2a}+1+\frac{3a+c}{2c}+1=\frac{3a+3c}{2a}+\frac{3a+3c}{2c}=\frac{3}{2}\left(a+c\right)\left(\frac{1}{a}+\frac{1}{c}\right)\)

\(M+2=\frac{3}{2}\left(1+\frac{a}{c}+\frac{c}{a}+1\right)=\frac{3}{2}\left(2+\frac{a}{c}+\frac{c}{a}\right)\)

Xét \(\frac{a}{c}+\frac{c}{a}\ge2\Leftrightarrow...\)(bạn tự biến đổi tương đương để chứng minh nó nhé)

(ĐK xảy ra dấu "=": a=c)

Do đó \(M+2=\frac{3}{2}\left(1+\frac{a}{c}+\frac{c}{a}+1\right)=\frac{3}{2}\left(2+\frac{a}{c}+\frac{c}{a}\right)\ge\frac{3}{2}\left(2+2\right)=6\Leftrightarrow M\ge4\)

Vậy GTNN của \(M=4\)khi \(a=c\Leftrightarrow\frac{2}{b}=\frac{2}{a}\Leftrightarrow b=a=c\)

Chúc bạn học tốt!

P/S: bài này khó thật đấy! Mình chuyên toán 9 mà giải hết nửa tiếng mới xong :D!

Ta có : \(3=ab+bc+ac\ge3\sqrt[3]{\left(abc\right)^2}\Rightarrow1\ge abc\)

\(\frac{bc}{a^2\left(b+2c\right)}+\frac{ac}{b^2\left(c+2a\right)}+\frac{ab}{c^2\left(a+2b\right)}\)

\(=\frac{\left(bc\right)^2}{abc\left(ab+2ac\right)}+\frac{\left(ac\right)^2}{abc\left(bc+2ab\right)}+\frac{\left(ab\right)^2}{abc\left(ca+2cb\right)}\)

\(\ge\frac{\left(ab+bc+ac\right)^2}{abc\left(3ab+3ac+3bc\right)}\)\(=\frac{3^2}{9abc}\)\(\ge1\)\(\left(dpcm\right)\)

a) \(\frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{c+a}=\frac{1}{5}\)

\(\Leftrightarrow\frac{2015}{a+b}+\frac{2015}{b+c}+\frac{2015}{c+a}=403\)

\(\Leftrightarrow\frac{a+b+c}{a+b}+\frac{a+b+c}{b+c}+\frac{a+b+c}{c+a}=403\)

\(\Leftrightarrow3+\frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{c+a}=403\)

\(\Leftrightarrow\frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{c+a}=400\)

b) \(\frac{a}{b}=\frac{c}{d}\Rightarrow\frac{a}{c}=\frac{b}{d}\)

Đặt \(\frac{a}{c}=\frac{b}{d}=k\Rightarrow\hept{\begin{cases}a=ck\\b=dk\end{cases}}\)

Thay vào rồi c/m nhé

a)  Từ đẳng thức : \(A=\frac{a}{b+c}+\frac{b}{a+c}+\frac{c}{a+b}\)

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

\(\Rightarrow A+3=\frac{a+b+c}{b+c}+\frac{a+b+c}{a+c}+\frac{a+b+c}{a+b}\)

\(\Rightarrow A+3=\left(a+b+c\right).\frac{1}{b+c}+\left(a+b+c\right).\frac{1}{a+c}+\left(a+b+c\right).\frac{1}{a+b}\)

\(\Rightarrow A+3=\left(a+b+c\right).\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)\)

\(\Rightarrow A+3=2015.\frac{1}{5}\)

\(\Rightarrow A+3=403\)

\(\Rightarrow A=400\)

Vậy A = 400

b) Đặt \(\frac{a}{b}=\frac{c}{d}=k\)

\(\Rightarrow a=bk;c=dk\)

Khi đó : \(\frac{2a^2-3ab+5b^2}{2a^2+3ab}=\frac{2\left(bk\right)^2-3b^2k+5b^2}{2\left(bk\right)^2+3b^2k}=\frac{2k^2b^2-3b^2k+5b^2}{2b^2k^2+3b^2k}=\frac{b^2\left(2k^2-3k+5\right)}{b^2\left(2k^2+3k\right)}\)

\(=\frac{2k^2-3k+5}{2k^2+3k}\left(1\right)\);

\(\frac{2c^2-3cd+5d^2}{2c^2+3cd}=\frac{2\left(dk\right)^2-3d^2k+5d^2}{2\left(dk\right)^2+3d^2k}=\frac{2d^2k^2-3d^2k+5d^2}{2d^2k^2+3d^2k}=\frac{d^2.\left(2k^2-3k+5\right)}{d^2\left(2k^2+3k\right)}\)

\(=\frac{2k^2-3k+5}{2k^2+3k}\left(2\right)\)

Từ (1) và (2) => \(\frac{2a^2-3ab+5b^2}{2a^2+3ab}=\frac{2c^2-3cd+5d^2}{2c^2+3cd}\)(đpcm)

ta có \(\frac{1}{a}+\frac{1}{c}=\frac{2}{b}\Rightarrow b=\frac{2ac}{a+c}\)

thay b vào\(\frac{a+b}{2a-b}+\frac{c+b}{2c-b}=\frac{a+3c}{2a}+\frac{c+3a}{2c}\)

                                                  \(=\frac{2ac+3\left(a^2+c^2\right)}{2ac}\ge\frac{2ac+6ac}{2ac}=4\)

a, Có: \(\frac{a}{b}=\frac{c}{d}\Leftrightarrow\frac{a}{c}=\frac{b}{d}\)

Áp dụng tính chất dãy tỉ số bằng nhau

\(\frac{a}{c}=\frac{b}{d}=\frac{2a}{2c}=\frac{3b}{3d}=\frac{2a-3b}{2c-3d}=\frac{2a+3b}{2c+3d}\)

Có: \(\frac{2a-3b}{2c-3d}=\frac{2a+3b}{2c+3d}\Leftrightarrow\frac{2a+3b}{2a-3b}=\frac{2c+3d}{2c-3d}\)

b, Co: \(\frac{a}{c}=\frac{b}{d}\Rightarrow\left(\frac{a}{c}\right)^2=\left(\frac{b}{d}\right)^2\Rightarrow\frac{ab}{cd}\)

Lại có:\(\left(\frac{a}{c}\right)^2=\left(\frac{b}{d}\right)^2\Rightarrow\frac{a^2}{c^2}=\frac{b^2}{d^2}\left(1\right)\)

Áp dụng tính chất dãy tỉ số bằng nhau

\(\frac{a^2}{c^2}=\frac{b^2}{d^2}=\frac{a^2-b^2}{c^2-d^2}\left(2\right)\)

Tu (1)&(2),có: \(\frac{ab}{cd}=\frac{a^2-b^2}{c^2-d^2}\)

Mình xem phép làm câu 1 ạ. 

Đề là?

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

Chứng minh tương đương 

\(\frac{a+b}{2a-b}+\frac{c+b}{2c-b}\ge4\)<=> 12ac - 9bc  - 9ab + 6b² \(\le\)0 ( quy đồng )  (2)

Từ (1) <=> 2ac = ab + bc  Thay vào (2) <=> 6ab + 6bc - 9bc  - 9ab + 6b²  \(\le\)0 

<=> a + c \(\ge\)2b 

Từ (1) => \(\frac{2}{b}=\frac{1}{a}+\frac{1}{c}\ge\frac{4}{a+c}\)

=> a + c \(\ge\)2b đúng => BĐT ban đầu đúng

Dấu "=" xảy ra <=> a = c = b