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? $\endgroup$ – Simply Beautiful Art May 2 '16 at 23:37
  • $\begingroup$ No, they can be any real numbers. $\endgroup$ – Karo May 3 '16 at 0:02
  • $\begingroup$ Oh, I'm sorry. You should have made the title more clear, lol, I solved the wrong problem. $\endgroup$ – Simply Beautiful Art May 3 '16 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. $\endgroup$ – Simply Beautiful Art May 3 '16 at 0:29

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 $z=x+y=0.$


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.