\(VT=\dfrac{2y+3z+5}{1+x}+1+\dfrac{3z+x+5}{2y+1}+1+\dfrac{x+2y+5}{1+3z}+1-3\)

\(VT=\dfrac{x+2y+3z+6}{1+x}+\dfrac{x+2y+3z+6}{1+2y}+\dfrac{x+2y+3z+6}{1+3z}-3\)

\(VT=24\left(\dfrac{1}{1+x}+\dfrac{1}{1+2y}+\dfrac{1}{1+3z}\right)-3\ge\dfrac{24.9}{1+x+1+2y+1+3z}-3=\dfrac{216}{21}-3=\dfrac{51}{7}\)

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

\(VT+3=\left(x+2y+3z+6\right)\left(\dfrac{1}{1+x}+\dfrac{1}{1+2y}+\dfrac{1}{1+3z}\right)\)

= \(24\left(\dfrac{1}{1+x}+\dfrac{1}{1+2y}+\dfrac{1}{1+3z}\right)\)

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

cộng 3 vào rồi b-c-s

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 : \(\frac{b+c+5}{a+1}+\frac{a+c+5}{b+1}+\frac{a+c+5}{c+1}=\frac{24-m}{m}+\frac{24-n}{n}+\frac{24-p}{p}\)

\(=24\left(\frac{1}{m}+\frac{1}{n}+\frac{1}{p}\right)-3\ge\frac{24.9}{m+n+p}-3=\frac{24.9}{\left(a+1\right)+\left(b+1\right)+\left(b+1\right)}-3\)

                                                       \(=\frac{24.9}{18+3}-3=\frac{51}{7}\)

\(a,\left(x^2-2x+3\right)\left(\dfrac{1}{2}x-5\right)\\ =x^2.\dfrac{1}{2}x-5x^2-2x.\dfrac{1}{2}x+2x.5+3.\dfrac{1}{2}x-15\\ =\dfrac{1}{2}x^3-5x^2-x^2+10x+\dfrac{3}{2}x-15\\ =\dfrac{1}{2}x^3-6x^2+\dfrac{23}{2}x-15\)

\(b,\left(x^2y^2-\dfrac{1}{3}xy+2y\right)\left(x-2y\right)\\ =x^3y-2x^2y^3-\dfrac{1}{3}x^2y+\dfrac{2}{3}xy^2+2xy-4y^2\)

a) \(\left(x^2-2x+3\right)\left(\dfrac{1}{2}x-5\right)\)

\(=\dfrac{1}{2}x^3-5x^2-x+10x+\dfrac{3}{2}x-15\)

\(=\dfrac{1}{2}x^3-5x^2+\dfrac{48}{5}x-15\)

b) \(\left(x^2y^2-\dfrac{1}{3}xy+2y\right)\left(x-2y\right)\)

\(=x^3y^2-2x^2y^3-\dfrac{1}{3}x^2y+\dfrac{2}{3}xy^2+2xy-4y^2\)

-\(\dfrac{1}{4}\)x²y

giúp m

\(\frac{2y+3z+5}{1+x}+1+\frac{3z+x+5}{1+2y}+1+\frac{x+2y+5}{1+3z}+1\ge\frac{51}{7}+3=\frac{72}{7}\left(1\right)\)

Vậy ta cần chứng minh Bđt (1) , ta có:

\(VT_{\left(1\right)}=\frac{2y+3z+6+x}{1+x}+\frac{3z+x+2y+6}{1+2y}+\frac{x+2y+3z+6}{1+3z}\)

\(=\left(3z+x+2y+6\right)\left(\frac{1}{1+x}+\frac{1}{1+2y}+\frac{1}{1+3z}\right)\)

Áp dụng Bđt \(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\ge\frac{9}{x+y+z}\)ta có:

\(\left(3z+x+2y+6\right)\left(\frac{1}{1+x}+\frac{1}{1+2y}+\frac{1}{3z}\right)\)

\(\ge\left(3z+x+2y+6\right)\left(\frac{9}{3+x+2y+3z}\right)\)

\(=\left(18+6\right)\cdot\frac{9}{18+3}=24\cdot\frac{3}{7}=\frac{72}{7}\)

Vậy Bđt (1) đúng =>Đpcm