Tinh tong : S= x+2y +3z, biet rang : \(\frac{1}{x+2y}+\frac{1}{2y+3z}+\frac{1}{3z+z}=\frac{12x}{2y+3z}+\frac{24y}{3z+x}-\frac{36z}{x+2y}=2016\)
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CB
3 tháng 6 2019
Đặt: \(\left\{{}\begin{matrix}x=a\\2y=b\\3z=c\end{matrix}\right.\Rightarrow a+b+c=18\)
Có: BDT
\(\Leftrightarrow\sum_{cyc}\left(\frac{b+c+5}{a+1}\right)\ge\frac{51}{7}\)
\(\Leftrightarrow\sum_{cyc}\left(\frac{a+b+c-a+5}{a+1}\right)\ge\frac{51}{7}\)(1)
Đặt tiếp tục: \(\left\{{}\begin{matrix}m=a+1\\n=b+1\\p=c+1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a=m-1\\b=n-1\\c=p-1\end{matrix}\right.\)
\(\left(1\right)\Leftrightarrow\sum_{cyc}\left(\frac{24-m}{m}\right)\ge\frac{51}{7}\)
\(\Leftrightarrow\sum_{cyc}\left(\frac{24}{m}-1\right)\ge\frac{51}{7}\)
\(\Leftrightarrow24\left(\frac{1}{m}+\frac{1}{n}+\frac{1}{p}\right)\ge\frac{72}{7}\)
\(\Leftrightarrow\frac{1}{m}+\frac{1}{n}+\frac{1}{p}\ge\frac{3}{7}\)
\(\Leftrightarrow\left(m+n+p\right)\left(\frac{1}{m}+\frac{1}{n}+\frac{1}{p}\right)\ge21\cdot\frac{3}{7}=9\)
\(\left(\frac{m}{n}-2+\frac{n}{m}\right)+\left(\frac{p}{m}-2+\frac{m}{p}\right)+\left(\frac{n}{p}-2+\frac{p}{n}\right)\ge0\)
\(\Leftrightarrow\frac{\left(m-n\right)^2}{mn}+\frac{\left(p-m\right)^2}{pm}+\frac{\left(n-p\right)^2}{pn}\ge0\)(đúng)
CB
4 tháng 6 2019
Đặt: \(\left\{{}\begin{matrix}x=a\\2y=b\\3z=c\end{matrix}\right.\)
BĐT
\(\Leftrightarrow\frac{b+c+5}{a+1}+\frac{a+c+5}{b+1}+\frac{a+b+5}{c+1}\ge\frac{51}{7}\)
\(\Leftrightarrow\frac{a+b+c-a+5}{a+1}+\frac{a+c+b-b+5}{b+1}+\frac{a+b+c-c+5}{c+1}\ge\frac{51}{7}\)
\(\Leftrightarrow\frac{24-\left(a+1\right)}{a+1}+\frac{24-\left(b+1\right)}{b+1}+\frac{24-\left(c+1\right)}{c+1}\ge\frac{51}{7}\)(1)
Đặt tiếp: \(\left\{{}\begin{matrix}a+1=m\\b+1=n\\c+1=p\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a=m-1\\b=n-1\\c=p-1\end{matrix}\right.\)
(1)\(\Leftrightarrow\frac{24-m}{m}+\frac{24-n}{n}+\frac{24-p}{p}\ge\frac{51}{7}\)
\(\Leftrightarrow24\left(\frac{1}{m}+\frac{1}{n}+\frac{1}{p}\right)-3\ge\frac{51}{7}\)
\(\Leftrightarrow24\left(\frac{1}{m}+\frac{1}{n}+\frac{1}{p}\right)\ge\frac{72}{7}\)
\(\Leftrightarrow\frac{1}{m}+\frac{1}{n}+\frac{1}{p}\ge\frac{3}{7}\)
\(\Leftrightarrow\left(m+n+p\right)\left(\frac{1}{m}+\frac{1}{n}+\frac{1}{p}\right)\ge\frac{3}{7}\left(m+n+p\right)\)( do m+n+p>0)
\(\Leftrightarrow3+\frac{m}{n}+\frac{n}{m}+\frac{p}{n}+\frac{n}{p}+\frac{m}{p}+\frac{p}{m}\ge\frac{3}{7}\left[\left(a+b+c\right)+3\right]\)
\(\Leftrightarrow\frac{m}{n}+\frac{n}{m}+\frac{p}{n}+\frac{n}{p}+\frac{p}{m}+\frac{m}{p}-6\ge0\)
Tới đây chắc bn làm đc rồi
27 tháng 10 2016
Đặt \(x+2y+3z=A\)
Áp dụng tính chất của dãy tỉ số bằng nhau có :
\(A=\frac{x+2y}{2y+3z-3}=\frac{2y+3z}{3z+x-3}=\frac{3z+x}{x+2y-3}=\frac{x+2y+2y+3z+3z+x}{x+2y+2y+3z+3z+x-3-3-3}\)
\(\Rightarrow A=\frac{2A}{2A-9}\)
\(\Rightarrow\frac{2}{2A-9}=1\)
\(\Rightarrow2A-9=2\)
\(\Rightarrow A=\frac{11}{2}\)
Cũng áp dụng tính chất của dãy tỉ số bằng nhau và có :
- \(A=\frac{x+2y}{2y+3z-3}=\frac{2y+3z}{3z+x-3}=\frac{3z+x}{x+2y-3}\)
\(=\frac{\left(x+2y\right)+\left(2y+3z\right)-\left(3z+x\right)}{\left(2y+3z-3\right)+\left(3z+x-3\right)-\left(x+2y-3\right)}=\frac{4y}{4y-3}=\frac{11}{2}\)
\(\Rightarrow2.\left(4y\right)=11.\left(4y-3\right)\)
\(\Rightarrow8y=44y-33\)
\(\Rightarrow36y=33\)
\(\Rightarrow y=\frac{11}{12}\)
- \(A=\frac{x+2y}{2y+3z-3}=\frac{2y+3z}{3z+x-3}=\frac{3z+x}{x+2y-3}\)
\(=\frac{\left(x+2y\right)-\left(2y+3z\right)+\left(3z+x\right)}{\left(2y+3z-3\right)-\left(3z+x-3\right)+\left(x+2y-3\right)}=\frac{2x}{2x-3}=\frac{11}{2}\)
\(\Rightarrow2.\left(2x\right)=11\left(2x-3\right)\)
\(\Rightarrow4x=22x-33\)
\(\Rightarrow18x=33\)
\(\Rightarrow x=\frac{33}{18}=\frac{11}{6}\)
\(\Rightarrow3z=A-x-2y=\frac{11}{2}-\frac{11}{6}-\frac{2.11}{12}=\frac{11}{6}\)
\(\Rightarrow z=\frac{11}{6}:3=\frac{11}{18}\)
Vậy ...
28 tháng 10 2016
Cho mình bổ sung : \(TH2:A=0\)
\(\Rightarrow2x=4y=6z=0\)
\(\Rightarrow x=y=z=0\)
Vậy ....
18 tháng 9 2016
Đặt \(\hept{\begin{cases}a=x\\b=2y\\c=3z\end{cases}}\) => a + b + c = 18
\(P=\frac{2y+3z+5}{1+x}+\frac{3z+x+5}{1+2y}+\frac{x+2y+5}{1+3z}=\frac{b+c+5}{a+1}+\frac{a+c+5}{b+1}+\frac{a+b+5}{c+1}\)
Lại đặt \(\hept{\begin{cases}m=a+1\\n=b+1\\p=c+1\end{cases}}\Rightarrow\hept{\begin{cases}a=m-1\\b=n-1\\c=p-1\end{cases}}\)
Ta có : \(\frac{b+c+5}{a+1}+\frac{a+c+5}{b+1}+\frac{a+c+5}{c+1}=\frac{24-m}{m}+\frac{24-n}{n}+\frac{24-p}{p}\)
\(=24\left(\frac{1}{m}+\frac{1}{n}+\frac{1}{p}\right)-3\ge\frac{24.9}{m+n+p}-3=\frac{24.9}{\left(a+1\right)+\left(b+1\right)+\left(b+1\right)}-3\)
\(=\frac{24.9}{18+3}-3=\frac{51}{7}\)
