Let $x$, $y$ be two positive numbers such that $x^4+y^4=x^2+y^2$. Prove that $$1\leqslant x+y\leqslant 2.$$ With $x+y\leqslant 2$. I tried

We have $$x^4+y^4\geqslant \dfrac{(x^2+y^2)^2}{2}.$$ Therefore, $$x^2+y^2\geqslant \dfrac{(x^2+y^2)^2}{2}$$ or $$(x^2+y^2)^2-2(x^2+y^2)\leqslant 0$$ Implies $x^2+y^2\leqslant 2.$ Another way $$\dfrac{x+y}{2} \leqslant \sqrt{\dfrac{x^2+y^2}{2}}=1.$$ Thus $x+y\leqslant 2.$

How can I prove $x+y\geqslant 1$?

Is this true?

Let $x$, $y$ be two positive numbers such that $x^m+y^m=x^n+y^n$, where $m$, $n$, ($m \neq n$) be two positive integer numbers, we have $$ x+y\leqslant 2.$$


If $x+y\lt 1$ then $(x+y)^2\lt 1$, and therefore $x^2+y^2\lt 1$.

But from $0\lt x^2+y^2\lt 1$, we conclude that $(x^2+y^2)^2 \lt x^2+y^2$, and therefore $x^4+y^4\lt x^2+y^2$.

  • $\begingroup$ you posted the same answer twice oO $\endgroup$ – Oussama Boussif Aug 8 '15 at 22:45
  • 1
    $\begingroup$ Yes, it has happened to me before, a strange bug. $\endgroup$ – André Nicolas Aug 8 '15 at 22:46

let $x+y=u,xy=v^2 \implies u^2 \ge 4v^2 \\ x^2+y^2=u^2-2v^2,x^4+y^4=(x^2+y^2)^2-2(xy)^2 =(u^2-2v^2)^2-2v^4=u^4-4u^2v^2+2v^4 \implies \\ u^4-(4v^2+1)u^2+2v^4+2v^2=0 \\ u^2=\dfrac{4v^2+1+ \sqrt{8v^4+1}}{2}$

$u^2=\dfrac{4v^2+1-\sqrt{8v^4+1}}{2} \le 2v^2$

$v^2\ge 0 \implies u^2=\dfrac{4v^2+1+ \sqrt{8v^4+1}}{2} \ge \dfrac{1+1}{2}=1$ when $v^2=0$


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