Ta thấy \(\sqrt{1+x^3}=\sqrt{\left(1+x\right)\left(1-x+2^2\right)}\) ≤\(\dfrac{x^2}{2}+2\)
\(P=\dfrac{x^2}{\sqrt{\left(1+x^3\right)\left(1+y^3\right)}}+\dfrac{y^2}{\sqrt{\left(1+y^3\right)\left(1+z^3\right)}}+\dfrac{z^2}{\left(\sqrt{1+z^3}\right)\left(1+x^3\right)}\) ≥\(\dfrac{4x^2}{\left(2+x^3\right)+\left(1+y^3\right)}+\dfrac{4y^2}{\sqrt{\left(1+y^3\right)\left(1+z^3\right).4}}+\dfrac{4z^2}{\sqrt{4\left(1+z^3\right)\left(2+x^2\right)}}\)
Đặt \(a=\dfrac{x^2}{4};b=\dfrac{y^2}{4};c=\dfrac{z^2}{4}\)
Khi đó x,y,z > 0 và xyz = 8 ⇒ a.b.c = 1(a,b,c > 0)
⇒ P ≥ \(\dfrac{16a}{\left(2+4a\right)\left(2+4b\right)}+\dfrac{16b}{\left(2+4b\right)\left(2+4c\right)}+\dfrac{16c}{\left(2+2c\right)\left(2+4a\right)}\)
(+) P ≥ \(4.\left[\dfrac{a}{\left(1+2a\right)\left(1+2b\right)}+\dfrac{b}{\left(1+2b\right)\left(1+2c\right)}+\dfrac{c}{\left(1+2c\right)\left(1+2a\right)}\right]\)
→ P ≥ \(4.\dfrac{a\left(1+2c\right)+b\left(1+2a\right)+c\left(1+2b\right)}{\left(1+2a\right)\left(1+2b\right)\left(1+2c\right)}\)
\(=4.\dfrac{a+b+c+2\left(ab+bc+ca\right)}{\left(1+2a\right)\left(1+2b\right)\left(1+2c\right)}=4.\dfrac{a+b+c+2\left(ab+bc+ca\right)}{1+2\left(a+b+c\right)+4.\left(1b+bc+ca\right)+8abc}\)
thấy a+b+c ≥ \(3\sqrt{abc}=3\) ⇒ a + b + c + 2(ab + bc + ca) ≥ 9
= 1+8abc
⇒ \(\left(1+8abc\right)+\left[2\left(a+b+c\right)+4\left(ab+bc+ca\right)\right]\) ≤ \(3\left[\left(a+b+c\right)+2\left(ab+bc+ca\right)\right]\)
⇒ P ≥ \(4\dfrac{a+b+c+2\left(ab+bc+ca\right)}{3\left[\left(a+b+c\right)+2\left(ab+bc+ca\right)\right]}=\dfrac{4}{3}\)
⇒ P ≥ \(\dfrac{4}{3}\) dấu '=' xảy ra ⇔ a = b = c = 1 ⇔ x = y = z = 2
