Find all triples of real numbers $x,y,z$ such that $1+x^4\leq 2(y-z)^2$, $1+y^4\leq 2(z-x)^2$, and $1+z^4\leq 2(x-y)^2$.

Beside $(1,0,-1)$ and permutations, I can't find any others. We cannot have two equal numbers. Maybe the inequality $1+a^4\geq 2a^2$ helps? It yields $|x-y|\geq |z|$ and analogous.

  • $\begingroup$ So $x,y,z$ are not required to be whole numbers? $\endgroup$ May 2, 2016 at 23:37
  • $\begingroup$ No, they can be any real numbers. $\endgroup$
    – user336268
    May 3, 2016 at 0:02
  • $\begingroup$ Oh, I'm sorry. You should have made the title more clear, lol, I solved the wrong problem. $\endgroup$ May 3, 2016 at 0:22
  • $\begingroup$ You would want $|x-y|\ge1,|y-z|\ge1,|z-x|\ge1$ so that the squared part is greater than $1$, so that it is at least larger than the LHS without the ^4 term. $\endgroup$ May 3, 2016 at 0:29

1 Answer 1


The only solutions are $(-1,0,1)$ and permutations.

Subtracting $2x^2,$ $2y^2,$ and $2z^2$ respectively from each inequality gives \begin{align*} (1-x^2)^2&=1+x^4-2x^2\leq 2(y-z)^2-2x^2&=&-2(x-y+z)(x+y-z)\\ (1-y^2)^2&=1+y^4-2y^2\leq 2(z-x)^2-2y^2&=&-2(x+y-z)(-x+y+z)\\ (1-z^2)^2&=1+z^4-2z^2\leq 2(x-y)^2-2z^2&=&-2(-x+y+z)(x-y+z). \end{align*} Since the left-hand-sides are non-negative, these inequalities can be multiplied: $$0\leq (1-x^2)^2(1-y^2)^2(1-z^2)^2\leq -8(-x+y+z)^2(x-y+z)^2(x+y-z)^2\leq 0.$$ So one of $-x+y+z,$ $x-y+z,$ or $x+y-z$ is zero. After a permutation of variables we have $x+y-z=0,$ which forces $(1-x^2)^2=(1-y^2)^2=0,$ so $x,y\in\{-1,1\}.$ We cannot have $x=y$ because $1\leq 1+z^4\leq 2(x-y)^2.$ So $x=-y\in\{-1,1\},$ which gives $z=x+y=0.$


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