abeker wrote: Determine the smallest possible real constant $C$ such that the inequality $$|x^3 + y^3 + z^3 + 1| \leq C|x^5 + y^5 + z^5 + 1|$$holds for all real numbers $x, y, z$ satisfying $x + y + z = -1$. For $x=y=z=-\frac{1}{3}$ we get $C\geq0.9$, but for $C=0.9$ we need to prove that $$|(x+y)(x+z)(y+z)|\sum_{cyv}(x^2-xy)\geq0.$$Done!

abeker wrote: Determine the smallest possible real constant $C$ such that the inequality $$|x^3 + y^3 + z^3 + 1| \leq C|x^5 + y^5 + z^5 + 1|$$holds for all real numbers $x, y, z$ satisfying $x + y + z = -1$. Let x, y and z real numbers such that $x+y+z=1$. Prove that: \[ |x^5+y^5+z^5-1| \geq \frac{10}{9}|x^3+y^3+z^3-1|\]

$$ x+y+z=-1$$$$xy+yz+zx=A$$$$xyz=B$$$$|x^3 + y^3 + z^3 + 1|=|3A+3B|$$$$x^5 + y^5 + z^5 + 1|=|(A+B)(5-5A)|$$$$(x+y+z)^2 \ge 3(xy+yz+zx)$$$$A\leq \frac{1}{3}$$$$2 \leq 3|A-1|$$For $x=y=z=-\frac{1}{3}$ we get $C\geq0.9$ $$3|A+B| \leq 0,9|A+B||5-5A|$$Answer:$0,9$