prove an inequality with conditions

Question

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

$a^2+b^2+c^2\le 1+2abc$ ?

Answer 1

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!

Answer 2

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.