Đặt \(a=x^2,b=y^3,c=z^3\)
+ \(\frac{a+b+c}{3}\ge\sqrt[3]{abc}\Leftrightarrow x^3+y^3+z^3\ge3xyz\)
\(\Leftrightarrow x^3+y^3+z^3-3xyz\ge0\)
\(\Leftrightarrow\left(x+y\right)^3+z^3-3xy\left(x+y\right)-3xyz\ge0\)
\(\Leftrightarrow\left(x+y+z\right)\left[\left(x+y\right)^2-\left(x+y\right)z+z^2\right]-3xy\left(x+y+z\right)\ge0\)
\(\Leftrightarrow\left(x+y+z\right)\left(x^2+2xy+y^2+z^2-yz-xz-3xy\right)\ge0\)
\(\Leftrightarrow\left(x+y+z\right)\left(x^2+y^2+z^2-xy-yz-zx\right)\ge0\)
\(\Leftrightarrow\frac{1}{2}\left(x+y+z\right)\left[\left(x-y\right)^2+\left(y-z\right)^2+\left(z-x\right)^2\right]\ge0\)
Vì BĐT cuối luôn đúng \(\forall x,y,z\ge0\) nên ta có đpcm
Dấu "=" \(\Leftrightarrow x=y=z\Leftrightarrow a=b=c\)
cách 2: Đặt \(a=x^3,b=y^3,c=z^3\)
Áp dụng BĐT Cauchy với 2 số ko âm :
\(\left(x^3+y^3\right)+\left(z^3+xyz\right)\ge2\sqrt{x^3y^3}+2\sqrt{xyz^4}=2\sqrt{xy}\left(xy+z^2\right)\)
Dấu "=" \(\Leftrightarrow x=y=z\)
+ \(xy+z^2\ge2\sqrt{xyz^2}=2\sqrt{xy}\cdot z\) Dấu "=" \(\Leftrightarrow z^2=xy\)
Do đó : \(x^3+y^3+z^3+xyz\ge2\sqrt{xy}\cdot2z\sqrt{xy}=4xyz\)
\(\Rightarrow x^3+y^3+z^3\ge3xyz\Rightarrow\frac{a+b+c}{3}\ge\sqrt[3]{abc}\)
Dấu "=" \(\Leftrightarrow x=y=z\Leftrightarrow a=b=c\)
