Không mất tính tổng quát, giả sử \(x\ge y\ge z\)

\(y^2-yz+z^2=y^2+\left(z-y\right)y\le y^2\Rightarrow\dfrac{1}{y^2-yz+z^2}\ge\dfrac{1}{y^2}\)

Tương tự: \(\dfrac{1}{z^2-xz+x^2}\ge\dfrac{1}{x^2}\)

\(\Rightarrow P\ge\dfrac{1}{x^2-xy+y^2}+\dfrac{1}{x^2}+\dfrac{1}{y^2}=\dfrac{1}{x^2-xy+y^2}+\dfrac{x^2-xy+y^2}{x^2y^2}+\dfrac{1}{xy}\)

\(P\ge2\sqrt{\dfrac{x^2-xy+y^2}{x^2y^2\left(x^2-xy+y^2\right)}}+\dfrac{1}{xy}=\dfrac{3}{xy}\ge\dfrac{12}{\left(x+y\right)^2}\ge\dfrac{12}{\left(x+y+z\right)^2}=3\)

Dấu "=" xảy ra khi \(\left(x;y;z\right)=\left(1;1;0\right)\) và hoán vị

\(\dfrac{\sqrt{1+x^3+y^3}}{xy}>=\sqrt{\dfrac{3}{xy}}\)

\(\dfrac{\sqrt{1+y^3+z^3}}{yz}>=\sqrt{\dfrac{3}{yz}}\)

\(\dfrac{\sqrt{1+z^3+x^3}}{xz}>=\sqrt{\dfrac{3}{xz}}\)

=>\(VT>=\sqrt{3}\left(\dfrac{1}{\sqrt{xy}}+\dfrac{1}{\sqrt{yz}}+\dfrac{1}{\sqrt{xz}}\right)=3\sqrt{3}\)

Gọi \(A=\sum\dfrac{x^3}{\sqrt{y^2+3}}\)

Theo Holder: \(A.A.\left(\left(y^2+3\right)+\left(z^2+3\right)+\left(x^2+3\right)\right)\ge\left(x^3+y^3+z^3\right)^3\)

\(\Rightarrow A^2\ge\dfrac{\left(x^3+y^3+z^3\right)^3}{x^2+y^2+z^2+9}\ge\dfrac{\left(x^3+y^3+z^3\right)^3}{x^2+y^2+z^2+3\left(xy+yz+zx\right)}=\dfrac{\left(x^3+y^3+z^3\right)^3}{\left(x+y+z\right)^2+xy+yz+zx}\ge\dfrac{\left(x^3+y^3+z^3\right)^3}{\left(x+y+z\right)^2+\dfrac{\left(x+y+z\right)^2}{3}}\)

Ta có đánh giá sau: \(x^3+y^3+z^3\ge\dfrac{\left(x^2+y^2+z^2\right)^2}{x+y+z}\ge\dfrac{\left(x+y+z\right)^3}{9}\)

\(\Rightarrow A^2\ge\dfrac{\dfrac{\left(x+y+z\right)^3}{9}}{\left(x+y+z\right)^2+\dfrac{\left(x+y+z\right)^2}{3}}=\dfrac{x+y+z}{12}\ge\dfrac{\sqrt{3\left(xy+yz+zx\right)}}{12}\ge\dfrac{1}{4}\)

\(\Rightarrow A\ge\dfrac{1}{2}\)

BĐT sai với $x=1,5; y=0,5; z=1$

\(\dfrac{x^2}{y^2}+\dfrac{y^2}{x^2}+4\ge3\left(\dfrac{x}{y}+\dfrac{y}{x}\right)\)

\(\Leftrightarrow2\left(\dfrac{x^2}{y^2}+\dfrac{y^2}{x^2}+4\right)\ge6\left(\dfrac{x}{y}+\dfrac{y}{x}\right)\)

\(\Leftrightarrow\dfrac{2x^2}{y^2}+\dfrac{2y^2}{x^2}+8\ge\dfrac{6x}{y}+\dfrac{6y}{x}\)

\(\Leftrightarrow\left(\dfrac{x^2}{y^2}+2+\dfrac{y^2}{x^2}\right)-4\left(\dfrac{x}{y}+\dfrac{y}{x}\right)+4+\dfrac{x^2}{y^2}-2.\dfrac{x}{y}+1+\dfrac{y^2}{x^2}-2.\dfrac{y}{x}+1\ge0\)

\(\Leftrightarrow\left(\dfrac{x}{y}+\dfrac{y}{x}\right)^2-4.\left(\dfrac{x}{y}+\dfrac{y}{x}\right)+4+\left(\dfrac{x}{y}-1\right)^2+\left(\dfrac{y}{x}-1\right)^2\ge0\)

\(\Leftrightarrow\left(\dfrac{x}{y}+\dfrac{y}{x}-2\right)^2+\left(\dfrac{x}{y}-1\right)^2+\left(\dfrac{y}{x}-1\right)^2\ge0^{\left(1\right)}\)

\(^{\left(1\right)}\)đúng \(\Rightarrowđpcm\)

Áp dụng BĐT : x⁴ + y⁴ ≥ 2x²y²

=> \(\dfrac{x^2}{y^2}+\dfrac{y^2}{x^2}\) ≥ 2 ( x , y > 0 )

TT , \(\dfrac{x}{y}+\dfrac{y}{x}\) ≥ 2 ( x , y > 0 )

Ta có : \(\dfrac{x^2}{y^2}+\dfrac{y^2}{x^2}\) + 4 ≥ 6 ( 1 )

\(3\left(\dfrac{x}{y}+\dfrac{y}{x}\right)\) ≥ 6 ( 2 )

Từ ( 1 ; 2) => đpcm

\(\dfrac{\sqrt{1+x^3+y^3}}{xy}\ge\dfrac{\sqrt{3\sqrt[3]{x^3y^3}}}{xy}=\dfrac{\sqrt{3xy}}{xy}=\dfrac{\sqrt{3}}{\sqrt{xy}}\)

Tương tự \(\dfrac{\sqrt{1+y^3+z^3}}{yz}\ge\dfrac{\sqrt{3}}{\sqrt{yz}};\dfrac{\sqrt{1+x^3+z^3}}{xz}\ge\dfrac{\sqrt{3}}{\sqrt{xz}}\)

\(\Rightarrow VT\ge\sqrt{3}\left(\dfrac{1}{\sqrt{xy}}+\dfrac{1}{\sqrt{yz}}+\dfrac{1}{\sqrt{xz}}\right)\ge\sqrt{3}.\dfrac{3}{\sqrt[3]{xyz}}=3\sqrt{3}\)

Dấu "=" xảy ra khi x=y=z=1

\(xy\ne0,x,y\ne1\)

\(A=\dfrac{x^{ }}{y^3-1}-\dfrac{y}{x^3-1}+\dfrac{2\left(x+y\right)}{x^2y^2+3}\)

\(xét:\dfrac{2\left(x+y\right)}{x^2y^2+3}=\dfrac{2}{x^2y^2+3}\left(1\right)\)

\(\dfrac{x^{ }}{y^3-1}-\dfrac{y}{x^3-1}=\dfrac{x^4-x-y^4+y}{\left(x^3-1\right)\left(y^3-1\right)}\left(2\right)\)

\(xét:\) \(x^4-x-y^4+y=\left(x-y\right)\left(x^3+x^2y+xy^2+y^3-1\right)\)

\(=\left(x-y\right)\left[\left(x+y\right)^3-3xy\left(x+y\right)+xy\left(x+y\right)-1\right]\)

\(=\left(x-y\right)\left(1-3xy+xy-1\right)\)

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

\(xét\) \(\left(y^3-1\right)\left(x^3-1\right)=x^3y^3-\left[\left(x+y\right)^3-3xy\left(x+y\right)\right]+1\)

\(=x^3y^3-\left(1-3xy\right)+1=x^3y^3+3xy=xy\left(x^2y^2+3\right)\)

\(\Rightarrow\left(2\right)\Leftrightarrow\dfrac{-2\left(x-y\right)}{x^2y^2+3}\)

\(\left(1\right)\left(2\right)\Rightarrow A=\dfrac{2}{x^2y^2+3}-\dfrac{2\left(x-y\right)}{x^2y^2+3}=\dfrac{2-2x+2y}{x^2y^2+3}\ne0\left(đề-sai\right)\)