Find all positive integers x , satisfying the inequality for $a,b,c \in \mathbb{R}^+ $

Question

Find all positive integers x such that for any $a,b,c \in \mathbb{R}^+ $ satisfying the inequality $$ x(ab+bc+ca) > 5(a^2+b^2+c^2) $$ and there must exist a triangle with a,b,c as its sides respectively.

I really don't know how to approach this question.Please help me.

Show us your trying. – Michael Rozenberg Aug 10 '18 at 15:00 — Michael Rozenberg, Aug 10 '18 at 15:00
This follows immediately from this MSE-question. – Dietrich Burde Aug 10 '18 at 15:02 — Dietrich Burde, Aug 10 '18 at 15:02

Batominovski · Answer 1 · 2018-08-10T18:28:47.153

1

Since $a$, $b$, and $c$ must form sides of a triangle, there exist $p,q,r>0$ such that $a=q+r$, $b=r+p$, and $c=p+q$. Hence, the required inequality is equivalent to $$x\,\big((p^2+q^2+r^2)+3(qr+rp+pq)\big)>10\,\big((p^2+q^2+r^2)+(qr+rp+pq)\big)$$ for all $p,q,r>0$.

By taking $p:=1$ and $q,r\to0^+$ in the inequality above, we need $x\geq 10$. On the other hand, if $x\geq 10$, then $$\begin{align}x\,\big((p^2+q^2+r^2)+3(qr+rp+pq)\big)&\geq 10\,\big((p^2+q^2+r^2)+3(qr+rp+pq)\big) \\ &> 10\,\big((p^2+q^2+r^2)+(qr+rp+pq)\big)\,.\end{align}$$ Thus, the set of all viable values of $x$ is $[10,\infty)$. If you demand that $x$ be an integer, then $x=10,11,12,\ldots$ are good values of $x$. The inequality is sharp if and only if $x=10$.

edited Aug 10 '18 at 18:28

answered Aug 10 '18 at 16:42

Batominovski

50,341

Dear @Batominovski $n=6$ is valid. Try $a=b=c.$ – Michael Rozenberg Aug 10 '18 at 17:44
@MichaelRozenberg Note that $x=6$ is certainly not valid for $(a,b,c)=(10,10,1)$. – Batominovski Aug 10 '18 at 18:27
We need existing only. See please better the problem. – Michael Rozenberg Aug 10 '18 at 18:28
What do you mean? The problem statement says that for any $a,b,c>0$ and there must exist a triangle with side lengths $a,b,c$. This is not a for-some scenario. We can argue a bit about the syntax, but I think this is what the OP meant. – Batominovski Aug 10 '18 at 18:29
The problem was changed! See the first post of the starting-poster. – Michael Rozenberg Aug 10 '18 at 18:51
Then I was right when I said that $x$ greater than $6$ was necessary but not sufficient. – Ataulfo Aug 10 '18 at 19:00

score 0 · Answer 2 · answered Aug 10 '18 at 15:03

0

With Cauchy-Schwarz $$ab+bc+ca\leq a^2+b^2+c^2$$ then $$5(a^2+b^2+c^2)< x(ab+bc+ca)\leq x(a^2+b^2+c^2)$$ then $x>5$ is an answer.

answered Aug 10 '18 at 15:03

Nosrati

30,522

But for $x=5$ we'll get a wrong inequality. – Michael Rozenberg Aug 10 '18 at 15:07

Cesareo · Answer 3 · 2018-08-10T16:28:28.427

With $M = \left( \begin{array}{ccc} 0 & \frac{1}{2} & \frac{1}{2} \\ \frac{1}{2} & 0 & \frac{1}{2} \\ \frac{1}{2} & \frac{1}{2} & 0 \\ \end{array} \right)$ and $v = (a,b,c)$ we have

$$ x v^{\top}M v - 5 v^{\top}I_3 v > 0 $$

or

$$ v^{\top}\left(x M - 5 I_3\right) v > 0 $$

so

$$ x M - 5 I_3 > 0 $$

should be positive definite with eigenvalues

$$ \left\{-\frac{1}{2} (x+10) > 0,-\frac{1}{2} (x+10) > 0,x-5 > 0\right\} $$

or finally

$$ (x < -10) \cap (x > 5) $$

which is a contradiction. Considering now that $a > 0, b > 0, c > 0$ we have that $x > 5$ suffices.

Ataulfo · Answer 4 · 2018-08-10T16:53:26.980

HINT AND QUESTION.- Never we have $ab+bc+ca\gt a^2+b^2+c^2$ so never we have $ab+bc+ca\gt n(a^2+b^2+c^2)$ for any $n\in\mathbb N$. It follows, as necessary condition, that the integer $x$ must be greater than $5$.

It remains to discuss the triangular condition. I can not see this clearly.

Look at the minimum value $x=6$. We have with the parameter $a$ that the couples $(b,c)=(x,y)$ that satisfy $$6(ax+ay+xy)\gt 5(a^2+x^2+y^2)$$ are all inside an ellipse. Making $a\ge b\ge c$ these ellipses they grow with $a$ but this $a$ can tend to zero and the corresponding ellipse tends to be a point. Here I do not see clearly the formation of a possible triangle. I leave the question here.

Find all positive integers x , satisfying the inequality for $a,b,c \in \mathbb{R}^+ $

4 Answers4