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
$$ x+y\leqslant 2.$$

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 $v^2=0$  
