Ta có:

\(M=\frac{2x+y}{xy}+\frac{3}{2x+y}=\frac{2x+y}{2}+\frac{3}{2x+y}\)

\(=\left(\frac{3}{8}.\frac{2x+y}{2}+\frac{3}{2x+y}\right)+\frac{5}{8}.\frac{2x+y}{2}\)

Có: \(\frac{3}{8}.\frac{2x+y}{2}+\frac{3}{2x+y}\ge2\sqrt{\frac{3}{8}.\frac{2x+y}{2}.\frac{3}{2x+y}}=\frac{3}{2}\)

Dấu '=' xảy ra <=> \(\frac{3}{8}.\frac{2x+y}{2}=\frac{3}{2x+y}\)

Có: \(\frac{5}{8}.\frac{2x+y}{2}\ge\frac{5}{8}\sqrt{2xy}=\frac{5}{4}\)

Dấu '=' xảy ra <=> 2x=y và xy=2

Do đó \(M\ge\frac{3}{2}+\frac{5}{4}=\frac{11}{4}\)

Dấu '=' xảy ra <=> x=1 và y=2

Vậy GTNN của  M là 11/4 khi x=1 và y=2

Áp dụng BĐT AM-GM ta có:

\(\frac{x^4}{y+3z}+\frac{y+3z}{16}+\frac{1}{4}+\frac{1}{4}\ge4\sqrt[4]{\frac{x^4}{y+3z}\cdot\frac{y+3z}{16}\cdot\frac{1}{4}\cdot\frac{1}{4}}=x\)

\(\Rightarrow\frac{x^4}{y+3z}\ge x-\frac{y+3z}{16}-\frac{1}{2}\).Tương tự ta có:

\(\frac{y^4}{z+3x}\ge y-\frac{z+3x}{16}-\frac{1}{2};\frac{z^4}{x+3y}\ge z-\frac{x+3y}{16}-\frac{1}{2}\)

Cộng theo vế ta có:

\(P\ge\frac{3}{4}\left(x+y+z\right)-\frac{3}{2}\ge\frac{3}{4}\cdot3-\frac{3}{2}=\frac{3}{4}\)

Dấu "=" khi x=y=z=1

xin cho mình hỏi sao x+y+z lại\(\ge\)xy+yz+zx vậy

\(P=2x+y+\frac{30}{x}+\frac{5}{y}\)

     \(=\frac{10x}{5}+\frac{5y}{5}+\frac{30}{x}+\frac{5}{y}\)

     \(=\frac{6x}{5}+\frac{4x}{5}+\frac{y}{5}+\frac{4y}{5}+\frac{30}{x}+\frac{5}{y}\)

      \(=\left(\frac{6x}{5}+\frac{30}{x}\right)+\left(\frac{4x}{5}+\frac{4y}{5}\right)+\left(\frac{y}{5}+\frac{5}{y}\right)\)

Áp dụng bất đẳng thức cô-si cho hai số không âm

\(\frac{6x}{5}+\frac{30}{x}\ge2\sqrt{\frac{6x}{5}.\frac{30}{x}}=2\sqrt{36}=2.6=12\) (1)

\(\frac{y}{5}+\frac{5}{y}\ge2\sqrt{\frac{y}{5}.\frac{5}{y}}=2\) (2)

Theo đề \(x+y\ge10\) suy ra

\(\frac{4x}{5}+\frac{4y}{5}=\frac{4\left(x+y\right)}{5}\ge\frac{4.10}{5}=8\) (2)

Cộng (1); (2) ; (3) vế theo vế ta được:

\(\frac{6x}{5}+\frac{30}{x}+\frac{y}{5}+\frac{5}{y}+\frac{4x}{5}+\frac{4y}{5}\ge12+2+8=22\)

Dấu "=" xảy ra <=> \(\hept{\begin{cases}\frac{6x}{5}=\frac{30}{x}\\\frac{y}{5}=\frac{5}{y}\end{cases}\Rightarrow\hept{\begin{cases}x^2=25\\y^2=25\end{cases}}}\)

Vì x;y dương nên (x;y) = (5;5)

\(P=2x+y+\frac{30}{x}+\frac{5}{y}\)

\(\Leftrightarrow P=0,8\left(x+y\right)+\left(1,2x+\frac{30}{x}\right)+\left(0,2y+\frac{5}{y}\right)\)

Áp dụng BĐT AM-GM ta có:

\(P\ge0,8\left(x+y\right)+2.\sqrt{1,2x.\frac{30}{x}}+2.\sqrt{0,2y.\frac{5}{y}}=8+12+2=22\)

Dấu " = " xảy ra <=> x=y=5

Vậy \(P_{min}=22\Leftrightarrow x=y=5\)

Dự đoán dấu "=" khi x = 2 ; y= 1

Áp dụng bđt Cô-si cho 3 số và bđt \(\frac{a^2}{m}+\frac{b^2}{n}\ge\frac{\left(a+b\right)^2}{m+n}\) ta được

\(P=2x^2+y^2+\frac{28}{x}+\frac{1}{y}\)

    \(=\left(\frac{7x^2}{4}+\frac{14}{x}+\frac{14}{x}\right)+\left(\frac{y^2}{2}+\frac{1}{2y}+\frac{1}{2y}\right)+\left(\frac{x^2}{4}+\frac{y^2}{2}\right)\)

    \(\ge3\sqrt[3]{\frac{7x^2.14.14}{4.x^2}}+3\sqrt[3]{\frac{y^2.1.1}{2.2y.2y}}+\frac{\left(x+y\right)^2}{4+2}\)

      \(=3.\sqrt[3]{\frac{7.14.14}{4}}+\frac{3}{\sqrt[3]{2^3}}+\frac{3^2}{6}=24\)

Dấu "=" khi x = 2 ; y = 1 

Bài toán easy!

\(P=\left(2x^2+8\right)+\left(y^2+1\right)+\frac{28}{x}+\frac{1}{y}-9\)

Áp dụng BĐT AM-GM,ta có:

\(P\ge8x+2y+\frac{28}{x}+\frac{1}{y}-9\)

\(=\left(7x+\frac{28}{x}\right)+\left(y+\frac{1}{y}\right)+\left(x+y\right)-9\)

\(\ge2\sqrt{7x.\frac{28}{x}}+2\sqrt{y.\frac{1}{y}}+\left(x+y\right)-9\)

\(\ge28+2+3-9=24\)

Dấu "=" xảy ra khi \(\hept{\begin{cases}2x^2=8\\y^2=1\end{cases}}\Leftrightarrow\hept{\begin{cases}x=2\\y=1\end{cases}}\)

Vậy \(P_{min}=24\Leftrightarrow\hept{\begin{cases}x=2\\y=1\end{cases}}\)

\(x\left(x-z\right)+y\left(y-z\right)=0\)\(\Leftrightarrow\)\(x^2+y^2=z\left(x+y\right)\)

\(\frac{x^3}{z^2+x^2}=x-\frac{z^2x}{z^2+x^2}\ge x-\frac{z^2x}{2zx}=x-\frac{z}{2}\)

\(\frac{y^3}{y^2+z^2}=y-\frac{yz^2}{y^2+z^2}\ge y-\frac{yz^2}{2yz}=y-\frac{z}{2}\)

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

Cộng lại ra minP=4, dấu "=" xảy ra khi \(x=y=z=1\)

\(\frac{4x + y}{xy} = \frac{4x}{xy} + \frac{y}{xy} = \frac{4}{y} + \frac{1}{x}\)

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

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

Theo BĐT Cauchy, ta có:

\(\frac{1}{x} + \frac{x}{4} \ge 2\sqrt{\frac{1}{x} \cdot \frac{x}{4}} = 2 \cdot \frac{1}{2} = 1\)

\(\frac{4}{y} + \frac{4y}{9} \ge 2\sqrt{\frac{4}{y} \cdot \frac{4y}{9}} = 2 \cdot \frac{4}{3} = \frac{8}{3}\)

\(P = \left( \frac{1}{x} + \frac{x}{4} \right) + \left( \frac{4}{y} + \frac{4y}{9} \right) + \left( \frac{x}{2} - \frac{x}{4} \right) - \frac{y}{4} - \frac{4y}{9}\)

\(=\left(\frac{1}{x}+\frac{x}{4}\right)+\left(\frac{4}{y}+\frac{4y}{9}\right)+\frac{x}{4}-\frac{25y}{36}\)

x+y=5

=>y=5-x

\(\frac{x}{4} - \frac{25y}{36} = \frac{5 - y}{4} - \frac{25y}{36} = \frac{5}{4} - \frac{y}{4} - \frac{25y}{36} = \frac{5}{4} - \frac{34y}{36} = \frac{5}{4} - \frac{17y}{18}\)

=>\(P \ge 1 + \frac{8}{3} + \frac{5}{4} - \frac{17y}{18}\)

Chọn điểm rơi tại y=3

Khi đó, ta sẽ có: \(P\ge1+\frac{8}{3}+\frac{5}{4}-\frac{17 \cdot3}{18}=\frac{25}{12}\)

Dấu '=' xảy ra khi y=3 và x=5-y=5-3=2

\(A=\frac{x}{x+1}+\frac{y}{y+1}+\frac{z}{z+1}\).Áp dụng BĐT Cauchy-Schwarz,ta có:

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

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

\(\ge3-\frac{9}{\left(x+y+z\right)+\left(1+1+1\right)}=\frac{3}{4}\)

Dấu "=" xảy ra khi x = y = z = 1/3

Vậy A min = 3/4 khi x=y=z=1/3

Bỏ chữ "Áp dụng bđt Cauchy-Schwarz,ta có:"giùm mình,nãy đánh nhầm ở bài làm trước mà quên xóa đi!