How can I prove remainder side of this inequality? 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 $x^4+y^4\lt x^2+y^2$.
