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The conjecture below is a modified version of this question:

Prove that, given a triangle with sides $a,b,c$, there exists a triangle with sides $a+2b,b+2c,c+2a$ that has an area three times the original

Conjecture:

If $u$ is the area of a triangle with sides $a,b,c$, and $v$ is the area of a triangle with sides $a+2b,b+2c,c+2a$, then ${\large{\frac{v}{u}}}\ge 9$.

Remarks:

  • For the equilateral case ($a=b=c$), we get ${\large{\frac{v}{u}}}=9$.
  • Limited data testing seems to support the truth of the conjecture.
  • Trying to prove the claim via Heron's formula appears to be a disaster.

Question:$\;$Is the conjecture true?

quasi
  • 58,772

1 Answers1

3

Let $x=b+c-a\geq 0$, $y=a+c-b\geq 0$, $z=a+b-c\geq 0$. Let $p=\frac{1}{2}(a+b+c)$ be the semiperimeter of the triangle with sides $a$, $b$, $c$. Similarly let $P=\frac{1}{2}(A+B+C)=3p$ be the semiperimeter of the triangle with sides $A=b+2c$, $B=c+2a$, $C=a+2b$. Then, by Heron's formula, the inequality is equivalent to $$ \begin{align}0&\leq \frac{v^2}{u^2}-9^2=\frac{P(P-A)(P-B)(P-C)}{p(p-a)(p-b)(p-c)}-9^2\\&= \frac{6((x^2y+y^2z+z^2x)+2(xy^2+yz^2+zx^2)-9xyz)}{xyz} \end{align}$$ which holds because by the AGM inequality $$x^2y+y^2z+z^2x\geq 3xyz\quad\mbox{and}\quad xy^2+yz^2+zx^2\geq 3xyz.$$

Robert Z
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