Bạn tham khảo:
Ta xét $x^{3}+y^3=(x+y)(x^2-xy+y^2)≥(x+y)xy$
⇔$x^{3}+y^3+1≥(x+y)xy+1$
⇔$x^{3}+y^3+1≥(x+y)xy+xyz$
⇔$x^{3}+y^3+1≥(x+y+z)xy$
⇔$\dfrac{1}{x^3+y^3+1}$ $\leq$ $\dfrac{1}{(x+y+z)xy}$
⇔$\dfrac{1}{x^3+y^3+1}$ $\leq$ $\dfrac{z}{x+y+z}$
Tương quan tự ta được
$\dfrac{1}{z^3+y^3+1}$ $\leq$ $\dfrac{x}{x+y+z}$
$\dfrac{1}{x^3+z^3+1}$ $\leq$ $\dfrac{y}{x+y+z}$
⇒$\dfrac{1}{x^3+y^3+1}+$ $\dfrac{1}{z^3+y^3+1}+$$\dfrac{1}{x^3+z^3+1}$$\leq$ $\dfrac{z}{x+y+z}+$ $\dfrac{x}{x+y+z}+$ $\leq$ $\dfrac{y}{x+y+z}=$\ $\dfrac{x+y+z}{x+y+z}=1$
Vậy $GTLN=1$
Học tốt
