Cho x, y, z là các số thực dương thỏa $x+y+z=1$.Tìm giá trị nhỏ nhất của biểu thức:
$P=\left ( 2x^{3}+\frac{3}{2}y^{2}-y+\frac{65}{54} \right )\left ( 2y^{3}+\frac{3}{2}z^{2}-z+\frac{65}{54} \right )\left ( 2z^{3}+\frac{3}{2}x^{2}-x+\frac{65}{54} \right )$
Ta có:
$$P=\left ( 2x^{3}+\frac{3}{2}y^{2}-y+\frac{65}{54} \right )\left ( 2y^{3}+\frac{3}{2}z^{2}-z+\frac{65}{54} \right )\left ( 2z^{3}+\frac{3}{2}x^{2}-x+\frac{65}{54} \right )$$
$$=\left ( 2x^{3}+\frac{3}{2}y^{2}-y+\frac{1}{6}+\frac{28}{27} \right )\left ( 2y^{3}+\frac{3}{2}z^{2}-z+\frac{1}{6}+\frac{28}{27} \right )\left ( 2z^{3}+\frac{3}{2}x^{2}-x+\frac{1}{6}+\frac{28}{27} \right )$$
$$=\left ( 2x^{3}+\frac{1}{6}(3z-1)^2+\frac{28}{27} \right )\left ( 2y^{3}+\frac{1}{6}(3y-1)^2+\frac{28}{27} \right )\left ( 2z^{3}+\frac{1}{6}(3x-1)^2+\frac{28}{27} \right )\geq 8(x^3+\frac{14}{27})(y^3+\frac{14}{27})(z^3+\frac{14}{27})$$
$$=\frac{8}{27^3}[(3x)^3+14][(3y)^3+14][(3z)^3+14]$$
Đặt $a = 3x; b=3y; c=3z$ thì $a+b+c = 3(x+y+z)=3$
Xét $A=[(3x)^3+14][(3y)^3+14][(3z)^3+14]=(a^3+14)(b^3+14)(c^3+14)$
Theo Cauchy: $a^3+a^3+1\geq 3a^2$
$b^3+b^3+1\geq 3b^2$
Suy ra $(a^3+14)(b^3+14\geq \frac{1}{4}(3a^2+\frac{27}{2})(3b^2+\frac{27}{2})=\frac{9}{4}(a^2+9)(b^2+9)=\frac{9}{4}(9a^2+9b^2+a^2b^2+81)\geq \frac{9}{4}(8(a^2+b^2)+a^2+b^2+2ab)\geq \frac{9}{4}(5(a+b)^2+80)=\frac{9}{4}(5(3-c)^2+80)$
Do đó: $A\geq \frac{9}{4}(5(3-c)^2+80)(c^3+14)=\frac{9}{4}(c-1)^2[c(c-2)^2+12c+50]+3375\geq 3375$
Suy ra $P=\frac{8}{27^3}A\geq \frac{1000}{27}$
Vậy $minP=\frac{1000}{27}$ đạt được khi $a=b=c=1$ hay $x=y=z=\frac{1}{3}$