\(VT=\frac{1}{xy}+\frac{1}{x^2+y^2}=\frac{1}{x^2+y^2}+\frac{1}{2xy}+\frac{1}{2xy}\)

Áp dụng BĐT Cauchy schawazr ta có :

\(\frac{1}{x^2+y^2}+\frac{1}{2xy}+\frac{1}{2xy}\ge\frac{\left(1+1\right)^2}{\left(x+y\right)^2}+\frac{1}{\frac{2\left(x+y\right)^2}{4}}=4+2=6\)

Vậy đẳng thức đã được chứng minh .

Dấu bằng xảy ra khi \(x=y=\frac{1}{2}\)

a: \(x^2+x+1=x^2+x+\dfrac{1}{4}+\dfrac{3}{4}=\left(x+\dfrac{1}{2}\right)^2+\dfrac{3}{4}>0\)

b: \(x-2\cdot\sqrt{x}\cdot\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{3}{4}=\left(\sqrt{x}-\dfrac{1}{2}\right)^2+\dfrac{3}{4}>0\)

c: \(=x^2-2\cdot x\cdot\dfrac{1}{2}y+\dfrac{1}{4}y^2+\dfrac{3}{4}y^2=\left(x-\dfrac{1}{2}y\right)^2+\dfrac{3}{4}y^2>0\forall x,y\ne0\)

Áp dụng BĐT AM-GM ta có:

\(\sqrt[3]{yz}\le\frac{y+z+1}{3}\Rightarrow\frac{x}{\sqrt[3]{yz}}\ge\frac{x}{\frac{y+z+1}{3}}=\frac{3x}{y+z+1}\)

Tương tự rồi cộng lại ta có:

\(VT\ge3\left(\frac{x}{y+z+1}+\frac{y}{x+z+1}+\frac{z}{x+y+1}\right)\)

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

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

\(=x^2+y^2+z^2\ge xy+yz+xz=VP\)

Đẳng thức xảy ra khi \(x=y=z=1\)

Áp dụng BĐT AM-GM ta có:

\sqrt[3]{yz}\le\frac{y+z+1}{3}\Rightarrow\frac{x}{\sqrt[3]{yz}}\ge\frac{x}{\frac{y+z+1}{3}}=\frac{3x}{y+z+1}3yz​≤3y+z+1​⇒3yz​x​≥3y+z+1​x​=y+z+13x​

Tương tự rồi cộng lại ta có:

VT\ge3\left(\frac{x}{y+z+1}+\frac{y}{x+z+1}+\frac{z}{x+y+1}\right)VT≥3(y+z+1x​+x+z+1y​+x+y+1z​)

=3\left(\frac{x^2}{xy+yz+x}+\frac{y^2}{xy+yz+y}+\frac{z^2}{yz+xz+z}\right)=3(xy+yz+xx2​+xy+yz+yy2​+yz+xz+zz2​)

\ge\frac{3\left(x^4+y^4+z^4\right)}{2\left(xy+yz+xz\right)+x+y+z}\ge\frac{\left(x^2+y^2+z^2\right)^2}{x^2+y^2+z^2}≥2(xy+yz+xz)+x+y+z3(x4+y4+z4)​≥x2+y2+z2(x2+y2+z2)2​

=x^2+y^2+z^2\ge xy+yz+xz=VP=x2+y2+z2≥xy+yz+xz=VP

Đẳng thức xảy ra khi x=y=z=1x=y=z=1

Áp dụng BĐT AM-GM cho 3 số không âm, ta có: \(0< \sqrt[3]{yz.1}\le\frac{y+z+1}{3}\Rightarrow\frac{x}{\sqrt[3]{yz}}\ge\frac{3x}{y+z+1}\)

Làm tương tự với 2 hạng tử còn lại rồi cộng theo vế thì có:

\(\frac{x}{\sqrt[3]{yz}}+\frac{y}{\sqrt[3]{zx}}+\frac{z}{\sqrt[3]{xy}}\ge3\left(\frac{x}{y+z+1}+\frac{y}{z+x+1}+\frac{z}{x+y+1}\right)\)

\(=3\left(\frac{x^2}{xy+xz+x}+\frac{y^2}{xy+yz+y}+\frac{z^2}{zx+yz+z}\right)\ge^{Schwartz}3.\frac{\left(x+y+z\right)^2}{x+y+z+2\left(xy+yz+zx\right)}\)

\(=3.\frac{x^2+y^2+z^2+2\left(xy+yz+zx\right)}{x+y+z+2\left(xy+yz+zx\right)}\ge9.\frac{xy+yz+zx}{\sqrt{3\left(x^2+y^2+z^2\right)}+2\left(x^2+y^2+z^2\right)}\)

\(=9.\frac{xy+yz+zx}{3+2.3}=xy+yz+zx\) => ĐPCM.

Dấu "=" xảy ra khi x=y=z=1.