1
$\begingroup$

Let $a$, $b$, $c$ be the sides of $\triangle ABC$.

Prove $$(abc)^2\geq\frac{4\Delta}{\sqrt{3}}$$ where $\Delta$ is the area of the triangle.

(Editor's note: As observed in the answers, the target relation is in error. It should be $$(abc)^2\geq\left(\frac{4\Delta}{\sqrt{3}}\right)^3$$ I incorporated the correction into the title. —@Blue)

  • Clearly, $a$, $b$, $c$ are the sides opposite to the angles $A$, $B$, $C$
  • I considered the point $F$ inside the triangle as the Fermat point of the triangle and named $FA=x$, $FB=y$, and $FC=z$.
  • Then $\angle AFB = \angle AFC = \angle BFC=120^\circ$, so we have $$a^2=y^2+z^2+yz$$ and similarly for $b^2$ and $c^2$.
  • Observe that
    $$xy+yz+zx=\frac{4\Delta}{\sqrt{3}}$$
  • So, we have a big inequality to prove: $$\left(x^2+xy+y^2\right)\left(y^2+yz+z^2\right)\left(z^2+zx+x^2\right)\geq xy+yz+zx$$ but I can't prove it.

Thanks for help

$\endgroup$
2
$\begingroup$

The actual inequality is $(abc)^2\ge \left(\dfrac{4\Delta}{\sqrt{3}}\right)^3$.I will show this one.

  • Lemma:

    $\displaystyle \sin{\alpha}+\sin{\beta}+\sin{\gamma} \le \frac{3\sqrt{3}}{2}$ when $\alpha$,$\beta$,$\gamma$ are angles of a triangle.Now we see that $\sin x$ is concave in $(0,\pi)$ so applying Jensen's inequality we get $\dfrac{\sin{\alpha}+\sin{\beta}+\sin{\gamma}}{3}\le \sin\left({\dfrac{\alpha+\beta+\gamma}{3}}\right)=\dfrac{\sqrt{3}}{2}$ So our lemma is proved.

Now to the actual problem.$a+b+c=2R(\sin{\alpha}+\sin{\beta}+\sin{\gamma})\le 3\sqrt{3}R$ (from the lemma). Now $$\dfrac{4\Delta}{\sqrt{3}}=\dfrac{abc}{R\sqrt{3}}\le \dfrac{3abc}{a+b+c}\le (abc)^{2/3}$$ where in the last step we have used AM-GM.Now cubing both sides we get the desired inequality.

$\endgroup$
3
$\begingroup$

Your inequality $(abc)^2 \ge \frac{4}{\sqrt 3}\triangle$ cannot be true in general, as a matter of simple dimensional analysis:

If you shrink the triangle by a linear factor of $2$, the left-hand side $(abc)^2$ becomes $64$ times smaller, whereas the right-hand side only becomes $4$ times smaller. Since both sides of the inequality are nonzero, even if it happens to be true to begin with, you can make it false by considering a small enough triangle similar to the original one.

$\endgroup$
0
$\begingroup$

I think you mean to the following: $$(abc)^{\frac{2}{3}}\geq\frac{4\Delta}{\sqrt3},$$ which is $$16\Delta^2\leq3\sqrt[3]{a^4b^4c^4}$$ or $$\sum_{cyc}(2a^2b^2-a^4)\leq\sqrt[3]{a^4b^4c^4}$$ or $$\sum_{cyc}\left(a^4-2a^2b^2+\sqrt[3]{a^4b^4c^4}\right)\geq0,$$ which is true by Schur and AM-GM: $$\sum_{cyc}\left(a^4-2a^2b^2+\sqrt[3]{a^4b^4c^4}\right)=\sum_{cyc}\left(\sqrt[3]{a^{12}}+\sqrt[3]{a^4b^4c^4}-2a^2b^2\right)\geq$$ $$\geq\sum_{cyc}\left(\sqrt[3]{a^9b^3}+\sqrt[3]{a^9c^3}-2a^2b^2\right)=\sum_{cyc}(a^3b+ab^3-2a^2b^2)\geq0.$$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.