Đặt \(x=a+b+2c;y=2a+b+c;z=a+b+3c\left(x,y,z>0\right)\)

Từ đó tính được: \(\hept{\begin{cases}a=z+y-2x\\b=5x-y-3z\\c=z-x\end{cases}}\)

Lúc đó \(A=\frac{4\left(z+y-2x\right)}{x}+\frac{\left(5x-y-3z\right)+3\left(z-x\right)}{y}-\frac{8\left(z-x\right)}{z}\)

\(=\frac{4z+4y}{x}-8+\frac{2x}{y}-1+\frac{8x}{z}-8\)

\(=\left(\frac{4y}{x}+\frac{2x}{y}\right)+\left(\frac{4z}{x}+\frac{8x}{z}\right)-17\)

\(\ge2\sqrt{\frac{4y}{x}.\frac{2x}{y}}+2\sqrt{\frac{4z}{x}.\frac{8x}{z}}-17=12\sqrt{2}-17\)(Theo BĐT Cô - si cho 2 số dương)

Đẳng thức xảy ra khi \(\hept{\begin{cases}\frac{4y}{x}=\frac{2x}{y}\\\frac{4z}{x}=\frac{8x}{z}\end{cases}}\Leftrightarrow\hept{\begin{cases}x=y\sqrt{2}\\z=x\sqrt{2}=2y\end{cases}}\Leftrightarrow\frac{z}{2}=\frac{x}{\sqrt{2}}=\frac{y}{1}\)

Đặt \(\frac{z}{2}=\frac{x}{\sqrt{2}}=\frac{y}{1}=k\left(k>0\right)\)thì \(\hept{\begin{cases}z=2k\\x=\sqrt{2}k\\y=k\end{cases}}\). Lúc đó \(\hept{\begin{cases}a=\left(3-2\sqrt{2}\right)k\\b=\left(5\sqrt{2}-7\right)k\\c=\left(2-\sqrt{2}\right)k\end{cases}}\)

Vậy \(MinA=12\sqrt{2}-17\), đạt được khi \(\hept{\begin{cases}a=\left(3-2\sqrt{2}\right)k\\b=\left(5\sqrt{2}-7\right)k\\c=\left(2-\sqrt{2}\right)k\end{cases}}\left(k>0\right)\)

Lời giải:
Áp dụng BĐT Am-Gm và Cauchy-Schwarz:

\(P+4=\frac{3b+3c}{2a}+\frac{4a+3c}{3b}+\frac{12b-12c}{2a+3c}+4=\frac{3b}{2a}+\frac{3c}{2a}+\frac{2a}{3b}+\frac{2a}{3b}+\frac{3c}{3b}+\frac{12b+8a}{2a+3c}\)

\(=(\frac{3b}{2a}+\frac{2a}{3b})+(\frac{3c}{2a}+\frac{3c}{3b})+(\frac{2a}{3b}+\frac{2a}{2a})+\frac{4(3b+2a)}{2a+3c}-1\)

\(\geq 2\sqrt{\frac{3b}{2a}.\frac{2a}{3b}}+3c.\frac{4}{2a+3b}+2a.\frac{4}{3b+2a}+\frac{4(3b+2a)}{2a+3c}-1\)

\(=2+\frac{4(3c+2a)}{2a+3b}+\frac{4(3b+2a)}{2a+3c}-1\geq 2+2\sqrt{\frac{4(3c+2a)}{2a+3b}.\frac{4(3b+2a)}{2a+3c}}-1\)

\(=2+8-1=9\)

\(\Rightarrow P\geq 5\)

Vậy $P_{\min}=5$

Bìa này muốn làm cân 2 bước nha 

Bước 1 ) CM BĐT \(\left(x+y+z\right)\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)\ge9\)

nó được CM như sau

áp dụng BĐT cô si ta đc 

\(\left(x+y+z\right)\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)\ge3.\sqrt[3]{xyz}.3\sqrt[3]{\frac{1}{x}.\frac{1}{y}.\frac{1}{z}}=9.\sqrt[3]{xyz.\frac{1}{x}.\frac{1}{y}.\frac{1}{z}}=9\)

dấu = xảy ra khi x=y=z

Bước 2 ) Theo CM bước 1 . áp dụng ta đc

\(\frac{ab}{a+3b+2c}=\frac{ab}{\left(a+c\right)+\left(b+c\right)+2b}=\frac{ab}{9}.\frac{9}{\left(a+c\right)+\left(b+c\right)+2b}\le\frac{ab}{9}.\left(\frac{1}{a+c}+\frac{1}{b+c}+\frac{1}{2b}\right)\)

CM tương tự ta đc

\(\frac{bc}{b+3c+2a}\le\frac{bc}{9}.\left(\frac{1}{a+c}+\frac{1}{a+b}+\frac{1}{2c}\right)\)

\(\frac{ca}{c+3a+2b}\le\frac{ca}{9}\left(\frac{1}{b+c}+\frac{1}{a+b}+\frac{1}{2a}\right)\)

cộng zế zới zế ta đc

\(A\le\frac{1}{9}\left(\frac{ab+bc}{a+c}+\frac{ab+ca}{b+c}+\frac{bc+ca}{a+b}+\frac{a}{2}+\frac{b}{2}+\frac{c}{2}\right)\)

\(A\le\frac{1}{9}\left(b+a+c+\frac{a+b+c}{2}\right)=\frac{a+b+c}{6}=\frac{6}{6}=1\)

=> MAx A=1 khi a=b=c=2

Ta có:

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

\(\le\frac{1}{16}.\left(\frac{1}{a+b}+\frac{1}{c+a}+\frac{2}{b+c}\right)\left(1\right)\)

Tương tự ta có: \(\hept{\begin{cases}\frac{1}{3a+2b+3c}\le\frac{1}{16}.\left(\frac{1}{b+c}+\frac{1}{a+b}+\frac{2}{c+a}\right)\left(2\right)\\\frac{1}{3a+3b+2c}\le\frac{1}{16}.\left(\frac{1}{c+a}+\frac{1}{b+c}+\frac{2}{a+b}\right)\left(3\right)\end{cases}}\)

Từ (1), (2), (3) \(\Rightarrow P\le\frac{1}{16}.\left(\frac{4}{a+b}+\frac{4}{b+c}+\frac{4}{c+a}\right)\)

\(=\frac{1}{4}.2017=\frac{2017}{4}\)

đề thi vào lớp 10 năm nay của tỉnh thanh hóa

Ta có:

sigma \(\frac{ab}{3a+4b+5c}=\) sigma \(\frac{2ab}{5\left(a+b+2c\right)+\left(a+3b\right)}\le\frac{2}{36}\left(sigma\frac{5ab}{a+b+2c}+sigma\frac{ab}{a+3b}\right)\)

Ta đi chứng minh: \(sigma\frac{ab}{a+b+2c}\le\frac{9}{4}\)

có: \(sigma\frac{ab}{a+b+2c}\le\frac{1}{4}\left(sigma\frac{ab}{c+a}+sigma\frac{ab}{b+c}\right)=\frac{1}{4}\left(a+b+c\right)=\frac{9}{4}\)

BĐT trên đúng nếu: \(sigma\frac{ab}{a+3b}\le\frac{9}{4}\)

Ta thấy: \(sigma\frac{ab}{a+3b}\le\frac{1}{16}\left(sigma\frac{ab}{a}+sigma\frac{3ab}{b}\right)=\frac{1}{16}\)( sigma \(b+sigma3a\)) \(=\frac{1}{4}\left(a+b+c\right)=\frac{9}{4}\)

\(\Leftrightarrow sigma\frac{ab}{3a+4b+5c}\le\frac{1}{18}\left(5.\frac{9}{4}+\frac{9}{4}\right)=\frac{3}{4}\)(1)

MÀ: \(\frac{1}{\sqrt{ab\left(a+2c\right)\left(b+2c\right)}}=\frac{2}{2\sqrt{\left(ab+2bc\right)\left(ab+2ca\right)}}\ge\frac{2}{2\left(ab+bc+ca\right)}\)

\(=\frac{3}{3\left(ab+bc+ca\right)}\ge\frac{3}{\left(a+b+c\right)^2}=\frac{3}{9^2}=\frac{1}{27}\)(2)

Từ (1) và (2) \(\Rightarrow T\le\frac{3}{4}-\frac{1}{27}=\frac{77}{108}\)

Vậy GTLN của biểu thức T là 77/108 <=> a=b=c=3

Ta có: BĐT phụ sau: \(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\ge\frac{9}{a+b+c}\)( CM bằng BĐT Shwars nha).Áp dụng ta có:

\(\frac{1}{a+3b+5c}+\frac{1}{b+3c+5a}+\frac{1}{3a+2b+4c}\ge\frac{9}{9a+6b+12c}=\frac{3}{3a+2b+4c}\left(1\right)\)

\(\frac{1}{b+3c+5a}+\frac{1}{c+3a+5b}+\frac{1}{3b+2c+4a}\ge\frac{9}{9b+6c+12a}=\frac{3}{3b+2c+4a}\left(2\right)\)

\(\frac{1}{c+3a+5b}+\frac{1}{a+3b+5c}+\frac{1}{3c+2a+4b}\ge\frac{9}{9c+6a+12b}=\frac{3}{3c+2a+4b}\left(3\right)\)

Cộng (1),(2) và (3) có:

\(2\left(\frac{1}{a+3b+5c}+\frac{1}{b+3c+5c}+\frac{1}{c+3a+5b}\right)+\left(\frac{1}{3a+2b+4c}+\frac{1}{3b+2c+4a}+\frac{1}{3c+2a+4b}\right)\ge3\left(\frac{1}{3a+2b+4c}+\frac{1}{3b+2c+4a}+\frac{1}{3c+2a+4b}\right)\)

\(\Rightarrow2VP\ge2VT\)

\(\RightarrowĐPCM\)