ĐKXĐ : \(x>\frac{1}{2};y>\frac{1}{2};z>\frac{1}{2}\)

Áp dụng ( a+b)² \(\ge4ab\)ta có : 

( x+ 2y)² = \(\left(\frac{2x+y}{2}+\frac{3y}{2}\right)^2\ge4.\left(\frac{2x+y}{2}\right).\frac{3y}{2}\)

\(\Rightarrow\left(x+2y\right)^2\ge3y\left(2x+y\right)\)

\(\Rightarrow\frac{2x+y}{x+2y}\le\frac{x+2y}{3y}\)

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

Tương tự : \(\frac{2y+z}{y\left(y+2\right)}\le\frac{1}{3}\left(\frac{2}{y}+\frac{1}{z}\right)\)

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

=> \(A\le\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\)

Ta có : \(\sqrt{\left(2x-1\right)1}\le\frac{2x-1+1}{2}\)

\(\Rightarrow\sqrt{2x-1}\le x\)

\(\Rightarrow\frac{1}{x}\le\frac{1}{\sqrt{2x-1}}\)

        \(\frac{1}{y}\le\frac{1}{\sqrt{2y-1}}\)

           \(\frac{1}{z}\le\frac{1}{\sqrt{2z-1}}\)

Do đó 

A \(\le\frac{1}{\sqrt{2x-1}}+\frac{1}{\sqrt{2y-1}}+\frac{1}{\sqrt{2z-1}}\)

Vậy Max A = 3 khi x = y = z = 1

Theo Cô-si ta có:

\(3=\frac{1}{\sqrt{2x-1}}+\frac{1}{\sqrt{2y-1}}+\frac{1}{\sqrt{2z-1}}\ge\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\)

\(\Rightarrow\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\le3\)

Xét:

\(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}-\Sigma_{cyc}\frac{2x+y}{x\left(x+2y\right)}=\frac{1}{3}\left[\frac{\left(x-y\right)^2}{xy\left(x+2y\right)}+\frac{\left(y-z\right)^2}{yz\left(y+2z\right)}+\frac{\left(z-x\right)^2}{zx\left(z+2x\right)}\right]\ge0\)

\(\Rightarrow\Sigma_{cyc}\frac{2x+y}{x\left(x+2y\right)}\le3\)

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

Tương tự:

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

Cộng lại:

\(16P\le4\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)=16\Rightarrow P\le1\)

dấu "=" xảy ra tại \(x=y=z=\frac{3}{4}\)

\(A=\frac{x}{x+1}+\frac{y}{y+1}+\frac{z}{z+1}\).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!

À mà để phải là tìm Max mới đúng chứ nhỉ?

Do đó,bạn sửa dòng: \(\ge3-\frac{9}{\left(x+y+z\right)+\left(1+1+1\right)}=\frac{3}{4}\) đến hết thành:

"\(\le3-\frac{9}{\left(x+y+z\right)+\left(1+1+1\right)}=\frac{3}{4}\)

Dấu "=" xảy ra khi x=y=z=1/3

Vậy A max = 3/4 khi x=y=z=1/3

\(ĐKXĐ:x,y,z\ge1\left(x,y,z\inℤ\right)\)

Ta có: \(\left(x+2y\right)^2=\left(\frac{2x+y}{2}+\frac{3y}{2}\right)^2\ge4.\frac{2x+y}{2}.\frac{3y}{2}=3y\left(2x+y\right)\)

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

Tương tự: \(\frac{2y+z}{y\left(y+2x\right)}\le\frac{1}{3}\left(\frac{2}{y}+\frac{1}{z}\right)\);\(\frac{2z+x}{z\left(z+2x\right)}\le\frac{1}{3}\left(\frac{2}{z}+\frac{1}{x}\right)\)

\(\Rightarrow A\le\frac{1}{3}.3\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)=\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\)(*)

Ta có: \(\sqrt{2x-1}=\sqrt{\left(2x-1\right).1}\le\frac{2x-1+1}{2}=x\)(BĐT Cô - si)

\(\Rightarrow\frac{1}{x}\le\frac{1}{\sqrt{2x-1}}\)

Tương tự: \(\frac{1}{y}\le\frac{1}{\sqrt{2y-1}}\);\(\frac{1}{z}\le\frac{1}{\sqrt{2z-1}}\)

\(\Rightarrow\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\le\frac{1}{\sqrt{2x-1}}+\frac{1}{\sqrt{2y-1}}+\frac{1}{\sqrt{2z-1}}=3\)(**)

Từ (*) và (**) suy ra \(A=\frac{2x+y}{x\left(x+2y\right)}+\frac{2y+z}{y\left(y+2z\right)}+\frac{2z+x}{z\left(z+2x\right)}\le3\)

Đẳng thức xảy ra khi x = y = z = 1

Từ đẳng thức đã cho suy ra \(x>\frac{1}{2};y>\frac{1}{2};z>\frac{1}{2}\)

Áp dụng\(\left(a+b\right)^2\ge4ab\)ta có \(\left(x+2y\right)^2=\left(\frac{2x+y}{2}+\frac{3y}{2}\right)^2\ge4\cdot\frac{2x+y}{2}\cdot\frac{3y}{2}\)

\(\Rightarrow\left(x+2y\right)^2\ge3y\left(2x+y\right)\)(Dấu "=" xảy ra <=> x=y)

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

Tương tự \(\hept{\begin{cases}\frac{2y+z}{y\left(y+2z\right)}\le\frac{1}{3}\left(\frac{2}{y}+\frac{1}{z}\right)\\\frac{2z+x}{z\left(z+2x\right)}\le\frac{1}{3}\left(\frac{2}{z}+\frac{1}{x}\right)\end{cases}}\)

=> \(A\le\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\)(Dấu "=" xảy ra <=> x=y=z)

Ta có \(\sqrt{\left(2x-1\right)\cdot1}\le\frac{\left(2x-1\right)+1}{2}\Rightarrow\sqrt{2x-1}\le x\Rightarrow\frac{1}{x}\le\frac{1}{\sqrt{2x-1}}\)

Tương tự \(\hept{\begin{cases}\frac{1}{y}\le\frac{1}{\sqrt{2y-1}}\\\frac{1}{z}\le\frac{1}{\sqrt{2z-1}}\end{cases}}\)

Do đó \(A\le\frac{1}{\sqrt{2x-1}}+\frac{1}{\sqrt{2y-1}}+\frac{1}{\sqrt{2z-1}}=3\)(dấu "=" xảy ra <=> x=y=z=1)

Vậy Max_A=3 đạt được khi x=y=z=1

áp dụng BĐT Cauchy ta có

\(\frac{x^3}{y+2z}+\frac{y+2z}{9}+\frac{1}{3}>=3\sqrt[3]{\frac{x^3}{y+2z}.\frac{\left(y+2z\right)}{9}.\frac{1}{3}}=x\)

\(=>\frac{x^3}{y+2z}>=x-\frac{y+2z}{9}-\frac{1}{3}\)

Tương tự \(\frac{y^3}{z+2x}>=y-\frac{z+2x}{9}-\frac{1}{3}\),\(\frac{z^3}{x+2y}>=z-\frac{x+2y}{9}-\frac{1}{3}\)

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

Mà x+y+z=3

\(=>P>=3-1-1=1\)

=>Min P=1 

Dấu "=" xảy ra khi x=y=z=1

bạn đăng bđt đi CTV,,,,mik lm vs

một cách khác khá hay nhưng dài hơn:

\(P=\frac{x^4}{xy+2xz}+\frac{y^4}{yz+2xy}+\frac{z^4}{xz+2yz}\ge\frac{\left(x^2+y^2+z^2\right)^2}{3\left(xy+yz+xz\right)}\ge\frac{x^2+y^2+z^2}{3}\ge\frac{\frac{1}{3}\left(x+y+z\right)^2}{3}=1\)

ap dung bdt \(x^{m+n}+y^{m+n}\ge x^my^n+x^ny^m\)  (bn tu cm )

\(\Rightarrow x^7+y^7=x^{3+4}+y^{3+4}\ge x^3y^4+x^4y^3\)

\(\Rightarrow\frac{x^2y^2}{x^2y^2+x^7+y^7}\le\frac{x^2y^2}{x^2y^2\left(1+xy^2+x^2y\right)}=\frac{1}{1+x^2y+y^2x}=\frac{1}{xyz+x^2y+y^2x}=\frac{1}{xy\left(x+y+z\right)}=\)

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

ttu \(P\le\frac{x+y+z}{x+y+z}=1\) đầu = xảy ra khi x=y=z=1

\(ĐK:x,y,z>\frac{1}{2}\)

Ta có: \(\left(x+2y\right)^2=\left(\frac{3y}{2}+\frac{y+2x}{2}\right)^2\ge4.\frac{3y}{2}.\frac{y+2x}{2}=3y\left(2x+y\right)\)\(\Rightarrow\frac{2x+y}{x+2y}\le\frac{x+2y}{3y}\Rightarrow\frac{2x+y}{x\left(x+2y\right)}\le\frac{x+2y}{3xy}=\frac{1}{3}\left(\frac{2}{x}+\frac{1}{y}\right)\)

Tương tự: \(\frac{2y+z}{y\left(y+2z\right)}\le\frac{1}{3}\left(\frac{2}{y}+\frac{1}{z}\right)\); \(\frac{2z+x}{z\left(z+2x\right)}\le\frac{1}{3}\left(\frac{2}{z}+\frac{1}{x}\right)\)

Cộng theo vế ba bất đẳng thức trên, ta được: \(VT\le\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\le\frac{1}{\sqrt{2x-1}}+\frac{1}{\sqrt{2y-1}}+\frac{1}{\sqrt{2z-1}}=3\)

Đẳng thức xảy ra khi x = y = z = 1

Từ giả thiết \(\Rightarrow x,y,z>\frac{1}{2}\)

Áp dụng \(\left(a+b\right)^2\ge4ab\) taoi có: 

\(\left(x+2y\right)^2=\left(\frac{2x+y}{2}+\frac{3y}{2}\right)^2\ge4\left(\frac{2x+y}{2}\right)\frac{3y}{2}\)

\(\Rightarrow\left(x+2y\right)^2\ge3y\left(2x+y\right)\)

Dấu '=' xảy ra <=> x=y

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

Tương tự: \(\frac{2y+z}{y\left(y+2z\right)}\le\frac{1}{3}\left(\frac{2}{y}+\frac{1}{z}\right),\frac{2z+x}{z\left(z+2x\right)}\le\frac{1}{3}\left(\frac{2}{z}+\frac{1}{x}\right)\)

\(\Rightarrow A\le\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\)

Dấu '=' xảy ra <=>x=y=z

Lại có: \(\sqrt{\left(2x-1\right)1}\le\frac{\left(2x-1\right)+1}{2}\Rightarrow\sqrt{\left(2x-1\right)}\le x\Rightarrow\frac{1}{x}\le\frac{1}{\sqrt{2x-1}}\)

Tương tự: \(\frac{1}{y}\le\frac{1}{\sqrt{2y-1}},\frac{1}{z}\le\frac{1}{\sqrt{2z-1}}\)

\(\Rightarrow A\le\frac{1}{\sqrt{2x-1}}+\frac{1}{\sqrt{2y-1}}+\frac{1}{\sqrt{2z-1}}=3\)

Dấu '=' xảy ra <=> x=y=z=1

Vậy GTLN của A=3 khi x=y=z=1

HSG toán 9 Quảng Nam năm 2018-2019

Giải: Từ đẳng thức đã cho suy ra: \(x>\frac{1}{2};y>\frac{1}{2};z>\frac{1}{2}\). Áp dụng (a+b)² >= 4ab ta có:

\(\left(x+2y\right)^2=\left(\frac{2x+y}{2}+\frac{3y}{2}\right)^2\ge4\cdot\left(\frac{2x+y}{2}\right)\cdot\frac{3y}{2}\)

\(\Rightarrow\left(x+2y\right)^2\ge3y\left(2x+y\right)\). Dấu "=" xảy ra <=> x=y

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

Tương tự \(\hept{\begin{cases}\frac{2y+z}{y\left(y+2z\right)}\le\frac{1}{3}\left(\frac{2}{y}+\frac{1}{z}\right)\\\frac{2z+x}{z\left(z+2x\right)}\le\frac{1}{3}\left(\frac{2}{z}+\frac{1}{x}\right)\end{cases}}\)

\(\Rightarrow A\le\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\left("="\Leftrightarrow x=y=z\right)\)

Ta có \(\sqrt{\left(2x-1\right)\cdot1}\le\frac{\left(2x-1\right)+1}{2}\Rightarrow\sqrt{2x-1}\le2\Rightarrow\frac{1}{x}\le\frac{1}{\sqrt{2x-1}}\)

Tương tự \(\frac{1}{y}\le\frac{1}{\sqrt{2y-1}},\frac{1}{z}\le\frac{1}{\sqrt{2z-1}}\)Do đó:

\(A\le\frac{1}{\sqrt{2x-1}}+\frac{1}{\sqrt{2y-1}}+\frac{1}{\sqrt{2z-1}}=3\)

Dấu "=" xảy ra <=> x=y=z=1

Vậy GTLN của A=3 đạt được khi x=y=z=1