Sửa đề: cho x,y,z dương. CMR \(\frac{x^3+y^3}{2xy}+\frac{y^3+z^3}{2yz}+\frac{z^3+x^2}{2xz}\ge x+y+z\)

Áp dụng BĐT AM-GM ta có:

\(x^3+y^3=\left(x+y\right)\left(x^2-xy+y^2\right)\)

\(\ge\left(x+y\right)\left(2\sqrt{x^2y^2}-xy\right)\)

\(=\left(x+y\right)\left(2xy-xy\right)=xy\left(x+y\right)\)

\(\Rightarrow\frac{x^3+y^3}{2xy}\ge\frac{xy\left(x+y\right)}{2xy}=\frac{x+y}{2}\)

Tương tự cho 2 BĐT còn lại ta có:

\(\frac{y^3+z^3}{2yz}\ge\frac{y+z}{2};\frac{z^3+x^3}{2xz}\ge\frac{x+z}{2}\)

Cộng theo vế 3 BĐT trên ta có:

\(VT\ge\frac{2\left(x+y+z\right)}{2}=x+y+z=VP\)

Đẳng thức xảy ra khi \(x=y=z\)

Đề sai rồi. Không cho x, y, z dương hay không là đã sai rồi. Giả sử đã cho dương rồi thì vẫn sai.

Thế \(x=y=z=2\) vào thì ta được

\(\frac{2^2+2^2}{2.2.2}+\frac{2^2+2^2}{2.2.2}+\frac{2^2+2^2}{2.2.2}\ge2+2+2\)

\(\Leftrightarrow3\ge6\) sai.

x³ + y³ = ( x + y ) ( x² - xy + y² )

( x + y ) \(\left(2\sqrt{x^2}y2-xy\right)\)

( x + y ) ( 2xy - xy ) = xy ( x + y )

\(x^3+\frac{y3}{2xv}>xy\left(x+\frac{x}{2xy}\right)=x+\frac{y}{2}\)

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\(y^3+\frac{z^3}{2yz}>y+\frac{z}{2}=z^3+\frac{x^3}{2x}>x+\frac{y}{2}\)

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\(VT>\frac{2\left(x+y+z\right)}{2}=x+y+z=VP\)

Ghi chú :Đẳng thức xảy ra khi x = v = z

a) x²+y²-4x+4y+8=0

⇔ (x-2)²+(y+2)²=0

\(\Leftrightarrow\left\{{}\begin{matrix}x-2=0\\y+2=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=2\\y=-2\end{matrix}\right.\)

b)5x²-4xy+y²=0

⇔ x²+(2x-y)²=0

\(\Leftrightarrow\left\{{}\begin{matrix}x=0\\2x-y=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=0\\y=0\end{matrix}\right.\)

c)x²+2y²+z²-2xy-2y-4z+5=0

⇔ (x-y)²+(y-1)²+(z-2)²=0

\(\Leftrightarrow\left\{{}\begin{matrix}x-y=0\\y-1=0\\z-2=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=y=1\\z=2\end{matrix}\right.\)

b: Ta có: \(5x^2-4xy+y^2=0\)

\(\Leftrightarrow x^2-\dfrac{4}{5}xy+y^2=0\)

\(\Leftrightarrow x^2-2\cdot x\cdot\dfrac{2}{5}y+\dfrac{4}{25}y^2+\dfrac{21}{25}y^2=0\)

\(\Leftrightarrow\left(x-\dfrac{2}{5}y\right)^2+\dfrac{21}{25}y^2=0\)

Dấu '=' xảy ra khi \(\left\{{}\begin{matrix}x=0\\y=0\end{matrix}\right.\)

d)3x²+3y²+3xy-3x+3y+3=0

⇔ 6x²+6y²+6xy-6x+6y+6=0

⇔ 3(x+y)²+3(x-1)²+3(y+1)²=0

\(\Leftrightarrow\left\{{}\begin{matrix}x+y=0\\x-1=0\\y+1=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=1\\y=-1\end{matrix}\right.\)

\(\frac{x^2+y^2+z^2-2xy+2xz-2yz}{x^2-2xy+y^2-z^2}\)

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

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

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

áp dụng bổ đề     \(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\ge\frac{9}{a+b+c}\)(bạn dùng cô-si,xét tích \(\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)\left(a+b+c\right)\))

\(\Leftrightarrow\frac{1}{x^2+2xy}+\frac{1}{y^2+2yz}+\frac{1}{z^2+2xz}\ge\frac{9}{\left(x+y+z\right)^2}=\frac{9}{1^2}\)

Hướng dẫn :\(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=0\Rightarrow\frac{xy+yz+zx}{xyz}=0\Rightarrow xy+yz+zx=0\)

Thay vào:\(x^2+2yz=x^2+yz+yz=x^2+yz-xy-zx=x\left(x-y\right)-z\left(x-y\right)=\left(x-y\right)\left(x-z\right)\)

Tương tự thay vào mà quy đồng