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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<a \leq b \leq c}. 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}.

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