P = 3x + 2y + 6/x + 8/y 
P = (3x/2 + 6/x) + (3x/2 + 3y/2) + (y/2 + 8/y) 
Ta có 3x/2 + 6/x >= 2.căn (3x/2.6/x) = 6 
dấu = xảy ra khi 3x/2 = 6/x <=> x = 2 
3x/2 + 3y/2 = 3/2.(x+y) >= 3/2.6 = 9 
dấu = xảy ra khi x + y = 6 
y/2 + 8/y >= 2.căn (y/2.8/y) = 4 
Dấu = xảy ra khi y/2 = 8/y <=> y = 4 
Vậy P >= 6 + 9 + 4 <=> P > = 19 
Dấu = xảy ra khi x = 2 và y = 4 
=> P min = 19

sai roi ban phai dung ca x+y>=6 nua chu

\(P=3x+2y+\frac{6}{x}+\frac{8}{y}\)

\(P=\left(\frac{3}{2}x+\frac{3}{2}y\right)+\left(\frac{3}{2}x+\frac{6}{x}\right)+\left(\frac{8}{y}+\frac{y}{2}\right)\)

\(P=\frac{3}{2}\left(x+y\right)+\left(\frac{3}{2}x+\frac{6}{x}\right)+\left(\frac{8}{y}+\frac{y}{2}\right)\)

\(\ge\frac{3}{2}.6+2\sqrt{\frac{3x}{2}.\frac{6}{x}}+2\sqrt{\frac{8}{y}.\frac{y}{2}}=9+6+4=19\)

\("="\Leftrightarrow x=2;y=4\)

các bạn biết ronaldo là ai không ?

\(P=3x+2y+\frac{6}{x}+\frac{8}{y}\)

\(2P=6x+4y+\frac{12}{x}+\frac{16}{y}\)

\(=\left(3x+\frac{12}{x}\right)+\left(y+\frac{16}{y}\right)+3\left(x+y\right)\)

\(\ge2\sqrt{3x\cdot\frac{12}{x}}+2\sqrt{y\cdot\frac{16}{y}}+3\cdot6=12+8+18=38\)( bđt AM-GM và giả thiết x + y ≥ 6 )

=> P ≥ 19

Đẳng thức xảy ra <=> \(\hept{\begin{cases}3x=\frac{12}{x}\\y=\frac{16}{y}\\x+y=6\end{cases}}\Rightarrow\hept{\begin{cases}x=2\\y=4\end{cases}}\)

Vậy MinP = 19

Ta có: \(P=3x+2y+\frac{6}{x}+\frac{8}{y}=\left(\frac{3}{2}x+\frac{3}{2}y\right)+\left(\frac{3}{2}x+\frac{6}{x}\right)+\left(\frac{y}{2}+\frac{8}{y}\right)\)

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

\(\frac{3x}{2}+\frac{6}{x}\ge2\sqrt{\frac{3x}{2}.\frac{6}{x}}=6;\frac{y}{2}+\frac{8}{y}\ge2\sqrt{\frac{y}{2}.\frac{8}{y}}=4\)

\(\Rightarrow P\ge9+6+4=19\)

Dấu '=' xảy ra <=> \(\hept{\begin{cases}x+y=6\\\frac{3x}{2}=\frac{6}{x}\\\frac{y}{2}=\frac{8}{y}\end{cases}}\Leftrightarrow\hept{\begin{cases}x=2\\y=4\end{cases}}\)

Vậy GTNN của P là 19

Ta có bđt \(\(\frac{1}{a}+\frac{1}{b}\ge\frac{4}{a+b}\)\)

\(\(\Rightarrow\frac{1}{a+b}\le\frac{1}{4}\left(\frac{1}{a}+\frac{1}{b}\right)\)\)

Áp dụng nhiều lần bđt trên ta được

\(\(\frac{1}{3x+3y+2z}=\frac{1}{\left(2x+y+z\right)+\left(x+2y+z\right)}\le\frac{1}{4}\left(\frac{1}{2x+y+z}+\frac{1}{x+2y+z}\right)\)\)

\(\(\le\frac{1}{4}\left(\frac{1}{\left(x+y\right)+\left(x+z\right)}+\frac{1}{\left(x+y\right)+\left(y+z\right)}\right)\)\)

\(\(\le\frac{1}{4}\left[\frac{1}{4}\left(\frac{1}{x+y}+\frac{1}{x+z}+\frac{1}{x+y}+\frac{1}{y+z}\right)\right]\)\)

\(\(\le\frac{1}{16}\left(\frac{2}{x+y}+\frac{1}{x+z}+\frac{1}{y+z}\right)\)\)

C/m tương tự cho các bđt còn lại

\(\(\frac{1}{3x+2y+3z}\le\frac{1}{16}\left(\frac{2}{x+z}+\frac{1}{x+y}+\frac{1}{y+z}\right)\)\)

\(\(\frac{1}{2x+3y+3z}\le\frac{1}{16}\left(\frac{2}{y+z}+\frac{1}{x+y}+\frac{1}{x+z}\right)\)\)

Cộng vế theo vế được

\(\(P\le\frac{1}{16}\left(\frac{4}{x+y}+\frac{4}{y+z}+\frac{4}{z+x}\right)=\frac{1}{4}\left(\frac{1}{x+y}+\frac{1}{y+z}+\frac{1}{z+x}\right)=\frac{1}{4}.6=\frac{3}{2}\)\)

Dấu "=" xảy ra

\(\(\Leftrightarrow\hept{\begin{cases}x=y=z\\\frac{1}{2x}+\frac{1}{2x}+\frac{1}{2x=6}\end{cases}}\)\)

\(\(\Leftrightarrow\hept{\begin{cases}x=y=z\\\frac{3}{2x}=6\end{cases}}\)\)

\(\(\Leftrightarrow\hept{\begin{cases}x=y=z\\x=\frac{1}{4}\end{cases}}\)\)

\(\(\Leftrightarrow x=y=z=\frac{1}{4}\)\)

Vậy ..........

cách khác :)) 

\(6=\frac{1}{x+y}+\frac{1}{y+z}+\frac{1}{z+x}\ge\frac{9}{2\left(x+y+z\right)}\)\(\Leftrightarrow\)\(x+y+z\le3\)

\(P=\frac{1}{3x+3y+2z}+\frac{1}{3x+2y+3z}+\frac{1}{2x+3y+3z}\)

\(P=\frac{1}{3\left(x+y+z\right)-z}+\frac{1}{3\left(x+y+z\right)-y}+\frac{1}{3\left(x+y+z\right)-x}\)

\(\ge\frac{9}{9\left(x+y+z\right)-\left(x+y+z\right)}=\frac{9}{8\left(x+y+z\right)}\ge\frac{9}{8.3}=\frac{3}{8}\)

Dấu "=" xảy ra \(\Leftrightarrow\)\(x=y=z=\frac{1}{4}\)

0000....78

a/ \(M=\left(x^2+\frac{1}{y^2}\right)\left(y^2+\frac{1}{x^2}\right)=x^2y^2+\frac{1}{x^2y^2}+2=\left(xy-\frac{1}{xy}\right)^2+4\ge4\)

Suy ra Min M = 4 . Dấu "=" xảy ra khi x=y=1/2

b/ Đề đúng phải là \(\frac{1}{3x+3y+2z}+\frac{1}{3x+2y+3z}+\frac{1}{2x+3y+3z}\ge\frac{3}{2}\)

Ta có \(6=\frac{1}{x+y}+\frac{1}{y+z}+\frac{1}{z+x}\ge\frac{9}{2\left(x+y+z\right)}\Rightarrow x+y+z\ge\frac{3}{4}\)

Lại có \(\frac{1}{3x+3y+2z}+\frac{1}{3x+2y+3z}+\frac{1}{2x+3y+3z}\ge\frac{9}{8\left(x+y+z\right)}\ge\frac{9}{8.\frac{3}{4}}=\frac{3}{2}\)

\(2A=6x+4y+\frac{12}{x}+\frac{16}{y}=3x+\frac{12}{x}+y+\frac{16}{y}+3x+3y\)

Áp dụng bất đẳng thức cô si cho 2 số dương, ta có:

\(3x+\frac{12}{x}\ge2.\sqrt{36}=12\)

\(y+\frac{16}{y}\ge2\sqrt{16}=8\)

Lại có\(x+y\ge6\Rightarrow3x+3y\ge18\)

Vậy \(2A\ge12+8+18\Leftrightarrow2A\ge38\Leftrightarrow A\ge19\)    \(a=19\Leftrightarrow x=2;y=4\)

Đặt \(x=a;2y=b;3z=c\Rightarrow a+b+c=3\) 

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

Áp dụng Bđt Cô si ngược dấu ta có:

\(T=\text{∑}a-\frac{a^2b}{1+b^2}\ge\text{∑}a-\frac{a^2b}{2b}=\text{∑}a-\frac{ab}{2}\)

\(=a+b+c-\frac{ab+bc+ca}{2}\ge a+b+c-\frac{\left(ab+bc+ca\right)^2}{6}\)\(=3-\frac{3^2}{6}=\frac{3}{2}\)

Dấu = khi \(a=b=c=1\Leftrightarrow\hept{\begin{cases}x=1\\y=\frac{1}{2}\\z=\frac{1}{3}\end{cases}}\)