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## Nice characterization side-lengths of a triangle

Find the greatest ${k}$ such that ${a,b,c>0}$ and ${kabc > a^3+b^3+c^3}$ implies that ${a,b,c}$ are the side-lengths of a triangle.

Solution: Note that if ${k>5}$ then for ${a=2,b=c=1}$ we have

$\displaystyle 2k=kabc >10=a^3+b^3+c^3$

and ${a,b,c}$ are not the side-lengths of a triangle. Therefore ${k \leq 5}$. We can assume without loss of generality that ${0. Pick ${k\leq 5}$ and suppose that ${a,b,c}$ are not the side-lengths of a triangle. Then ${c \geq a+b}$ and we obtain

$\displaystyle 5ab \geq kab > c^2+\frac{a^3+b^3}{c} \geq c^2+\frac{(a+b)^3}{4c}.$

We have used the fact that ${4(a^3+b^3) \geq (a+b)^3}$ (this comes from the fact that the power function is convex).

Consider now the function ${\displaystyle f_d: x \mapsto x^2+\frac{d^3}{4x}}$. Then ${\displaystyle f_d'(x)=2x-\frac{d^3}{4x^2}=\frac{(2x)^3-d^3}{x^2}}$ and we can see that for ${x \geq d/2}$ ${f_d}$ is strictly increasing. Therefore the above inequalities imply that

$\displaystyle 5ab > f_{a+b}(c) \geq f_{a+b}(a+b)=\frac{5}{4}(a+b)^2$

which is a contradiction. Therefore, any ${k \leq 5}$ makes the statement true, so the greatest ${k}$ is ${5}$.

An interesting challenge is to generalize this. One could try to modify the power ${3}$ in the RHS to an arbitrary power ${p \geq 4}$: the problem turns to: find the least ${k}$ such that ${kabc > a^p+b^p+c^p}$ implies that ${a,b,c}$ are the sides of a triangle. At first sight we see that this cannot be true, since the above inequality cannot imply that ${a,b,c}$ are sides of a triangle since if we scale ${a,b,c}$ with the same dilation parameter, the RHS converges to zero with an order greater than the LHS, so at one point the inequality will hold, no matter who the initial ${a,b,c}$ are. Still, one can reproduce the above machinery and use the ${2,1,1}$ example to see that ${k \leq 2^{p-1}+1}$, and then for each ${k \leq 2^{p-1}+1}$ if the inequality holds we obtain for ${a \leq b \leq c}$ and ${c \geq a+b}$ that

$\displaystyle ab > \left( \frac{a+b}{2}\right)^{p-1},$

but this is not anymore a contradiction for ${p>3}$. One immediate way to fix this is to assume ${a,b,c}$ are positive integers.

We have thus proved that the greatest ${k}$ such that ${a,b,c \in \Bbb{N}^*}$ and ${kabc>a^p+b^p+c^p}$ implies that ${a,b,c}$ are the side-lengths of a triangle is ${k=2^{p-1}+1}$.