Cho ba số thực dương x,y,z thoả mãn :x+2y+3z=18 .Chứng minh rằng :
\(\dfrac{2y+3z+5}{1+x}+\dfrac{3z+x+5}{1+2y}+\dfrac{x+2y+5}{1+3z}\ge\dfrac{51}{7}\)
Cho ba số thực x,y,z thoả mãn : x+2y+3z=18
Cmr : \(\dfrac{2y+3z+5}{1+x}+\dfrac{3z+x+5}{1+2y}+\dfrac{x+2y+5}{1+3z}\ge\dfrac{51}{7}\)
\(VT=\dfrac{2y+3z+5}{1+x}+1+\dfrac{3z+x+5}{2y+1}+1+\dfrac{x+2y+5}{1+3z}+1-3\)
\(VT=\dfrac{x+2y+3z+6}{1+x}+\dfrac{x+2y+3z+6}{1+2y}+\dfrac{x+2y+3z+6}{1+3z}-3\)
\(VT=24\left(\dfrac{1}{1+x}+\dfrac{1}{1+2y}+\dfrac{1}{1+3z}\right)-3\ge\dfrac{24.9}{1+x+1+2y+1+3z}-3=\dfrac{216}{21}-3=\dfrac{51}{7}\)
cho các số thực dương x,y,z thỏa x+2y+3z=18 CMR
\(\dfrac{2y+3z+5}{1+x}+\dfrac{3z+x+5}{1+2y}+\dfrac{x+2y+5}{1+3z}\ge\dfrac{51}{7}\)
\(VT+3=\left(x+2y+3z+6\right)\left(\dfrac{1}{1+x}+\dfrac{1}{1+2y}+\dfrac{1}{1+3z}\right)\)
= \(24\left(\dfrac{1}{1+x}+\dfrac{1}{1+2y}+\dfrac{1}{1+3z}\right)\)
Áp dụng BĐT cauchy-schwarz:
\(\dfrac{1}{1+x}+\dfrac{1}{1+2y}+\dfrac{1}{1+3z}\ge\dfrac{9}{3+x+2y+3z}=\dfrac{9}{21}\)
\(\Rightarrow VT\ge\dfrac{24.9}{21}-3=\dfrac{51}{7}\)
dấu = xảy ra khi x=2y=3z=6 hay x=6,y=3,z=2
cho x,y,z là các số dương thoả mãn \(\dfrac{1}{x+y}+\dfrac{1}{y+z}+\dfrac{1}{z+x}\)=6
Chứng minh \(\dfrac{1}{3x+3y+2z}+\dfrac{1}{3x+2y+3z}+\dfrac{1}{2x+3y+3z}\)≤\(\dfrac{3}{2}\)
Áp dụng BĐT Cauchy-Schwarz:
\(\dfrac{1}{x+y}+\dfrac{1}{x+y}+\dfrac{1}{y+z}+\dfrac{1}{z+x}\ge\dfrac{16}{3x+3y+2z}\\ \Leftrightarrow\dfrac{1}{3x+2y+2z}\le\dfrac{1}{16}\left(\dfrac{2}{x+y}+\dfrac{1}{y+z}+\dfrac{1}{z+x}\right)\\ \Leftrightarrow\sum\dfrac{1}{3x+2y+2z}\le\dfrac{1}{16}\left(\dfrac{4}{x+y}+\dfrac{4}{y+z}+\dfrac{4}{z+x}\right)=\dfrac{4}{16}\cdot6=\dfrac{3}{2}\)
Dấu \("="\Leftrightarrow x=y=z=\dfrac{1}{3}\)
cho x,y,z là các số thực dương thỏa mãn \(x^2+y^2+z^2\ge\dfrac{1}{3}\)
chứng minh \(\dfrac{x^3}{2x+3y+5z}+\dfrac{y^3}{2y+3z+5x}+\dfrac{z^3}{2z+3x+5y}\ge\dfrac{1}{30}\)
đặt\(A=\dfrac{x^3}{2x+3y+5z}+\dfrac{y^3}{2y+3z+5x}+\dfrac{z^3}{2z+3x+5y}\)
\(=>A=\dfrac{x^4}{2x^2+3xy+5xz}+\dfrac{y^4}{2y^2+3yz+5xy}+\dfrac{z^4}{2z^2+3xz+5yz}\)
BBDT AM-GM
\(=>A\ge\dfrac{\left(x^2+y^2+z^2\right)^2}{2\left(x^2+y^2+z^2\right)+8\left(xy+yz+xz\right)}\)
theo BDT AM -GM ta chứng minh được \(xy+yz+xz\le x^2+y^2+z^2\)
vì \(x^2+y^2\ge2xy\)
\(y^2+z^2\ge2yz\)
\(x^2+z^2\ge2xz\)
\(=>2\left(x^2+y^2+z^2\right)\ge2\left(xy+yz+xz\right)< =>xy+yz+xz\le x^2+y^2+z^2\)
\(=>2\left(x^2+y^2+z^2\right)+8\left(xy+yz+xz\right)\le10\left(x^2+y^2+z^2\right)\)
\(=>A\ge\dfrac{\left(x^2+y^2+z^2\right)^2}{10\left(x^2+y^2+z^2\right)}=\dfrac{x^2+y^2+z^2}{10}=\dfrac{\dfrac{1}{3}}{10}=\dfrac{1}{30}\left(đpcm\right)\)
dấu"=" xảy ra<=>x=y=z=1/3
Cho x+2y+3z=18; x,y,z là các số dương. CMR:
\(\frac{2y+3z+5}{1+x}+\frac{3z+x+5}{1+2y}+\frac{x+2y+5}{1+3z}\ge\frac{51}{7}\)
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)
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
Cho các số thực dương x,y,z thỏa x+2y+3y=18
CMR: \(\frac{2x+3y+5}{1+x}+\frac{3z+x+5}{1+2y}+\frac{x+2y+5}{1+3z}\ge\frac{51}{7}\)
1) cho ba số thực dương x,y,z thõa mãn : x + 2y +3z = 2
Tìm giá trị lớn nhất của biểu thức :
S = \(\sqrt{\dfrac{xy}{xy+3z}}+\sqrt{\dfrac{3yz}{3yz+x}}+\sqrt{\dfrac{3xz}{3xz+4y}}\)
Cho x, y, z>0. Chứng minh rằng:
\(\dfrac{x}{x+2y+3z}+\dfrac{y}{y+2z+3x}+\dfrac{z}{z+2x+3y}\ge\dfrac{1}{2}\)
\(VT=\dfrac{x^2}{x^2+2xy+3zx}+\dfrac{y^2}{y^2+2yz+3xy}+\dfrac{z^2}{z^2+2zx+3yz}\)
\(VT\ge\dfrac{\left(x+y+z\right)^2}{x^2+y^2+z^2+5xy+5yz+5zx}=\dfrac{\left(x+y+z\right)^2}{\left(x+y+z\right)^2+3\left(xy+yz+zx\right)}\ge\dfrac{\left(x+y+z\right)^2}{\left(x+y+z\right)^2+\left(x+y+z\right)^2}=\dfrac{1}{2}\)
cho x,y,z là các số thực dương thỏa mãn x+y+z=xyz.CMR
\(\dfrac{x}{1+x^2}+\dfrac{2y}{1+y^2}+\dfrac{3z}{1+z^2}=\dfrac{xyz\left(5x+4y+3z\right)}{\left(x+y\right)\left(y+z\right)\left(z+x\right)}\)