Just notice the equation is invariant if you flip $c$ over $a+b$. It follows if you assume $a+b+c$ minimal and then WLOG $a \le b \le c$ that $a+b \ge c$ so we're done.

Please anyone could give me the full solution? Thank you very much

calhanSPheiro2 wrote: Please anyone could give me the full solution? Thank you very much What's wrong with the solution that dinboy gave?

In equation: $2ab+2ac+2bc-a^2-b^2-c^2=k$ If you choose a number $t$ so the number $a$ was intact. Then decisions can be recorded. $a=\frac{k+t^2}{4}$ $b=\frac{k+t^2}{4}+t+1$ $c=k+(t+1)^2$

Consider only numbers $(a,b,c,k)$ that satisfy the equation and domain. WLOG $a=\max\lbrace a,b,c\rbrace$. Then our equation is $(b+c-a)^2=4bc-k$, so $4bc-k^2$ is a perfect square. What's more $x^2-2x(b+c)+(b-c)^2+k=0$ and $-2(b+c)<0\wedge (b-c)^2+k>0$ so both roots in $x$ are positive. Hence $b+c-\sqrt{4bc-k}>0$. Because this is an integer we can take $a=b+c-\sqrt{4bc-k}$. Then $a+b>c\wedge a+c>b$ as $a$ is maximal and $b,c>0$. Moreover $a=b+c-\sqrt{4bc-k}<b+c$ so there exists a non-degenerate triangle with sides $(a,b,c)$.