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
  • 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


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.


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.


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: $$\sum_{cyc}\left(a^4-2a^2b^2+\sqrt[3]{a^4b^4c^4}\right)=\sum_{cyc}\left(\sqrt[3]{a^{12}}+\sqrt[3]{a^4b^4c^4}-2a^2b^2\right)\geq$$ $$\geq\sum_{cyc}\left(\sqrt[3]{a^9b^3}+\sqrt[3]{a^9c^3}-2a^2b^2\right)=\sum_{cyc}(a^3b+ab^3-2a^2b^2)\geq0.$$


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