<s>[INST] 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 
$a^2+b^2+c^2\le 1+2abc$ ?
 [/INST] 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!
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