Mongolian Mathematical Olympiad

For any $x\le y$, we have $\sqrt{y-x}\le \frac{y-x+1}{2}$. Thus it suffices to prove that $$ab(b-a+1)+bc(c-b+1)+ca(c-a+1)=ab(2b+c)+bc(2c+a)+ca(2c+b)<\frac{1}{2}.$$ Since $(a+b+c{)}^3=a^3+b^3+c^3+3(a{b}^2+b{c}^2+a{c}^2)+3(a^{2}b+b^{2}c+a^{2}c)+6abc$, this is equivalent to proving the following: $$a{b}^2+a{c}^2+b{c}^2<a^3+b^3+c^3+3(a^{2}b+a^{2}c+b^{2}c).$$ Here $a{b}^2\le b^3$ and $b{c}^2+a{c}^2\le 2b{c}^2\le c^3+b^{2}c$ implies that $a{b}^2+a{c}^2+b{c}^2\le b^3+c^3+b^{2}c$. Since it is impossible to have $a^3=b^{2}c=c^3=0$ and $a+b+c=1$, inequality never holds.