6-th Czech-Slovak Match

Let positive numbers $a,b,c$ satisfy the inequality $5abc>a^3+b^3+c^3$. Let us show that there exists a triangle with the sides $a,b,c$. On the contrary, suppose that there is no such triangle. Then for $a,b,c$ at least one of the triangle inequalities is not valid. Let, e.g. $c\ge a+b$, i.e. $c=a+b+x$, where $x\ge 0$. Then from the initial inequality it follows that $$5ab(a+b+x)>a^3+b^3+(a+b{)}^3+3(a+b{)}^{2}x+e(a+b)x^2+x^3$$ or $$2a^{2}b+2a{b}^2>2a^3+2b^3+(a+b{)}^3+abx+3(a^2+b^2)x+3(a+b)x^2+x^3$$ Since the last four summands at the left side are nonnegative, we have $$\begin{array}{rlr}2a^{2}b+2a{b}^2& >2a^3+2b^3& \iff \\ a^3+b^3& -a^{2}b-a{b}^2<0& \iff \\ (a+b)(a-b{)}^2& <0& \end{array}$$ which is impossible. □

In fact, assume the contrary: Assume that there are three positive reals $a,b,c$ satisfying $5abc>a^3+b^3+c^3$, but they are not the sidelengths of a triangle. Since everything is symmetric, we can WLOG assume that $a\ge b\ge c$. Then, $c+a>b$ (since $a\ge b$) and $a+b>c$ (since $a\ge c$), so that we must have $b+c\le a$ (else, the positive reals $a,b,c$ would be the sidelengths of a triangle). Hence, $c+a-b>0$, $a+b-c>0$ and $b+c-a\le 0$, so that $$(b+c-a)(c+a-b)(a+b-c)\le 0$$ Thus, $$5abc-(b+c-a)(c+a-b)(a+b-c)\ge 5abc$$ On the other hand, $5abc>a^3+b^3+c^3$. Thus, $$\begin{array}{rlr}5abc-(b+c-a)(c+a-b)(a+b-c)& >a^3+b^3+c^3& \iff \\ 10abc+(a^3+b^3+c^3)-(a+b+c)(bc+ca+ab)& >a^3+b^3+c^3& \iff \\ 10abc& >(a+b+c)(bc+ca+ab)& \iff \end{array}$$ Division by $abc$ yields $$10>(a+b+c)\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)$$ Now, since $10=3^2+1$, the IMO 2004 problem 4 yields that $a,b,c$ are the sidelengths of a triangle, contradicting our assumption. Thus, our assumption was wrong, and we are done. ☐