Inequality with $x,y,z\geq 0$, $x+y+z=1.$ With $x,y,z\geq 0$, $x+y+z=1$.Prove that $$\sqrt{x+y^2}+\sqrt{y+z^2}+\sqrt{z+x^2}\geq 2 \tag{i}$$
The hint is using a lemma: 
If $a,b,c,d\geq 0 $satisfying $a+b=c+d$ and$|a-b|\leqslant|c-d|$ then we have $\sqrt{a}+\sqrt{b}\geq \sqrt{c}+\sqrt{d}$
How to prove this lemma? And is there a different way to prove the inequality (i)?

There is the following Vo Quoc Ba Can's solution.
We need to prove that 
$$\sum\limits_{cyc}\left(\sqrt{x+y^2}-y\right)\geq1$$ or
$$\sum\limits_{cyc}\frac{x}{\sqrt{x+y^2}+y}\geq1.$$
Now, by AM-GM 
$$\sum\limits_{cyc}\frac{x}{\sqrt{x+y^2}+y}=\sum\limits_{cyc}\frac{x(x+y)}{(x+y)\sqrt{x+y^2}+y(x+y)}\geq\sum\limits_{cyc}\frac{x(x+y)}{\frac{1}{2}((x+y)^2+x+y^2)+y(x+y)}.$$
Thus, it remains to prove that 
$$\sum\limits_{cyc}\frac{x(x+y)}{2x^2+4y^2+5xy+xz}\geq\frac{1}{2},$$
which is $$\sum\limits_{cyc}(4x^4y^2+16x^4yz+3x^3y^2z-19x^3z^2y-4x^2y^2z^2)\geq0,$$
which is obvious. 
Done!
