prove an inequality with conditions Let $a,b, c \in \mathbb{R}$  such that $0\le a\le1,0\le b\le1 , 0\le c\le1.$
If $$a+b\leq c+1,   
 \\ a+c \leq b+1,  
  \\b+c\leq a +1$$
can we prove that 
$a^2+b^2+c^2\le 1+2abc$ ?
 A: Let $a+1-b-c=x$, $b+1-a-c=y$ and $c+1-a-b=z$.
Hence, $x$, $y$ and $z$ are non-negatives such that $x+y+z=3-a-b-c\leq3$,
$a=\frac{2-y-z}{2}$, $b=\frac{2-x-z}{2}$, $c=\frac{2-x-y}{2}$ 
and we need to prove that
$$\frac{1}{4}\sum_{cyc}(2-x-y)^2\leq1+\frac{1}{4}\prod_{cyc}(2-x-y)$$ or
$$\sum_{cyc}(4+2x^2-8x+2xy)\leq4+8-\prod_{cyc}(x+y)+\sum_{cyc}(-8x+2x^2+6xy)$$ or
$$4(xy+xz+yz)\geq(x+y)(x+z)(y+z)$$
and since $1\geq\frac{x+y+z}{3}$, it's enough to prove that
$$\frac{4}{3}(x+y+z)(xy+xz+yz)\geq(x+y)(x+z)(y+z)$$ or
$$\sum_{cyc}(x^2y+x^2z+2xyz)\geq0.$$
Done!
A: Let
$$ a+b=(c+1) k_1,a+c=(b+1) k_2,b+c=(a+1) k_3 $$
and then
$$ a=-\frac{-k_2+k_3-k_1 \left(k_2 \left(k_3+2\right)+1\right)}{k_2+k_3+k_1 \left(1-k_2
   k_3\right)+2},b=-\frac{k_2-k_3-k_1 \left(\left(k_2+2\right) k_3+1\right)}{k_2+k_3+k_1
   \left(1-k_2 k_3\right)+2},c= -\frac{-k_3-k_2 \left(2 k_3+1\right)+k_1 \left(1-k_2 k_3\right)}{k_2+k_3+k_1 \left(1-k_2
   k_3\right)+2}. $$
Here $0\le k_1,k_2,k_3\le 1$. An easy calculation shows 
\begin{eqnarray}
&&a^2+b^2+c^2-1-2abc\\
&=&-\frac{4 (k_1+1)^2(k_2+1)^2(k_3+1)(2-k_1-k_2-k_3+k_1k_2 k_3)}{(k_2+k_3+k_1(1-k_2 k_3)+2)^3}.
\end{eqnarray}
Noting
\begin{eqnarray}
&&2-k_1-k_2-k_3+k_1k_2 k_3\\
&=&(1-k_1)+1-k_2-k_3+(k_1-1)k_2k_3+k_2k_3\\
&=&(1-k_1)(1-k_2k_3)+(1-k_2)(1-k_3)\ge0
\end{eqnarray}
one has
$$ a^2+b^2+c^2-1-2abc\le 0$$
or
$$ a^2+b^2+c^2\le1+2abc. $$
The job is done.
