# The isoperimetric inequality for triangles

If a triangle has perimeter $p$ and area $T$, how to prove that $p^2 \ge 12\sqrt{3} T$ and $T \le \frac{\sqrt{3}}{4}\cdot (abc)^{\frac{2}{3}}$?

I try some AM-GM inequality but I am stuck on it.

For the second one, write $T^3 = \dfrac{ab\sin C}{2}\cdot \dfrac{bc\sin A}{2}\cdot \dfrac{ca\sin B}{2}\implies T = \sqrt[3]{\dfrac{(abc)^2\sin A\sin B\sin C}{8}}$. Thus you are left with "proving" that $\dfrac{\sqrt[3]{\sin A\sin B\sin C}}{2} \le \dfrac{\sqrt{3}}{4}$. You have $\dfrac{\sqrt[3]{\sin A\sin B\sin C}}{2} \le \dfrac{1}{2}\cdot \dfrac{\sin A+\sin B+\sin C}{3}\le \dfrac{1}{2}\cdot \dfrac{3\sin\left(\dfrac{A+B+C}{3}\right)}{3}=\dfrac{1}{2}\cdot \sin\left(\dfrac{\pi}{3}\right) = \dfrac{\sqrt{3}}{4}$

Back to the first one, draw a diagram of such a triangle, and see for yourself that $p^2 \ge 12\sqrt{3}T \iff (2s)^2 \ge 12\sqrt{3}sr \iff s \ge 3\sqrt{3}r \iff \dfrac{s}{r} \ge 3\sqrt{3}$.

Your task now is to calculate $\dfrac{s}{r} = \dfrac{a+b+c}{2r}= \dfrac{1}{2}\cdot \dfrac{a+b+c}{r} = \dfrac{1}{2}\cdot \left(\dfrac{a}{r}+\dfrac{b}{r}+\dfrac{c}{r}\right)$ $= \dfrac{1}{2}\cdot \left(\left(\cot(B/2)+\cot(C/2)\right)+\left(\cot(C/2)+\cot(A/2)\right)+\left(\cot(A/2)+\cot(B/2)\right)\right)= \cot(A/2)+\cot(B/2)+\cot(C/2) \ge 3\cot\left(\dfrac{A+B+C}{6}\right) = 3\cot({\pi}/{6}) = 3\sqrt{3}$

by Jensen's inequality. Done!

• Is it Jensen's inequality instead of AM-GM inequality? – Ray Cheng Jun 17 '17 at 6:00
• AM-GM first, then Jensen ( = concavity ) after. – DeepSea Jun 17 '17 at 6:01
• Could you show more details please? I cannot proceed right now. – Ray Cheng Jun 17 '17 at 6:04
• will do ! ......in a minute... – DeepSea Jun 17 '17 at 6:06
• no need to draw diagram. already proven – Ray Cheng Jun 17 '17 at 6:17

Let $s=p/2$. Then $T^2=s(s-a)(s-b)(s-c)$. If we write $x=s-a$, $y=s-b$ and $z=s-c$ then $x>0$, $y>0$, $z>0$ and $s=x+y+z$. By AM/GM $s^3\ge27 xyz$. Your first inequality should follow.

The first inequality.

We need to prove that $$(a+b+c)^2\geq3\sqrt{3(a+b+c)\prod_{cyc}(a+b-c)}$$ or $$(a+b+c)^3\geq27\prod_{cyc}(a+b-c)$$ or $$\left(\frac{\sum\limits_{cyc}(a+b-c)}{3}\right)^3\geq\prod_{cyc}(a+b-c)$$ or $$\frac{\sum\limits_{cyc}(a+b-c)}{3}\geq\sqrt[3]{\prod_{cyc}(a+b-c)},$$ which is AM-GM.

The second inequality.

We need to prove that: $$\frac{1}{4}\sqrt{\sum_{cyc}(2a^2b^2-a^4)}\leq\frac{\sqrt3}{4}\sqrt[3]{a^2b^2c^2}$$ or $$a^4+b^4+c^4+3\sqrt[3]{a^4b^4c^4}\geq2(a^2b^2+a^2c^2+b^2c^2),$$ which follows from Schur and Muirhead: $$\sum_{cyc}(a^4-2a^2b^2+\sqrt[3]{a^4b^4c^4})=$$ $$=\sum_{cyc}(a^4-\sqrt[3]{a^8b^4}-\sqrt[3]{a^8c^4}+\sqrt[3]{a^4b^4c^4})+\sum_{cyc}(\sqrt[3]{a^8b^4}+\sqrt[3]{a^8c^4}-2a^2b^2)\geq0.$$ Done!

The Schur's inequality is the following.

Let $x$, $y$ and $z$ be positive numbers. Prove that: $$x^3+y^3+z^3- x^2y-x^2z-y^2x-y^2z-z^2x-z^2y+3xyz\geq0.$$

Proof.

Since the last inequality is homogeneous, we can assume that $x\geq y\geq z$ and we obtain: $$x^3+y^3+z^3- x^2y-x^2z-y^2x-y^2z-z^2x-z^2y+3xyz=$$ $$=x(x-y)(x-z)+y(y-x)(y-z)+z(z-x)(z-y)\geq$$ $$\geq x(x-y)(x-z)+y(y-x)(y-z)=(x-y)^2(x+y-z)\geq0.$$

Let $a^4=x^3$, $b^4=y^3$ and $c^4=z^3$.

Hence by Schur, $$a^4+b^4+c^4+3\sqrt[3]{a^4b^4c^4}-2(a^2b^2+a^2c^2+b^2c^2)=$$ $$=x^3+y^3+z^3+3xyz-2\left(\sqrt{x^3y^3}+\sqrt{x^3z^3}+\sqrt{y^3z^3}\right)=$$ $$=x^3+y^3+z^3-x^2y-x^2z-y^2x-y^2z-z^2x-z^2y+3xyz+$$ $$+x^2y+x^2z+y^2x+y^2z+z^2x+z^2y-2\left(\sqrt{x^3y^3}+\sqrt{x^3z^3}+\sqrt{y^3z^3}\right)\geq$$ $$\geq x^2y+x^2z+y^2x+y^2z+z^2x+z^2y-2\left(\sqrt{x^3y^3}+\sqrt{x^3z^3}+\sqrt{y^3z^3}\right)=$$ $$=xy(\sqrt{x}-\sqrt{y})^2+xz(\sqrt{x}-\sqrt{z})^2+yz(\sqrt{y}-\sqrt{z})^2\geq0.$$

• What is Schur and Muirhead? – Ray Cheng Jun 17 '17 at 5:24
• @Ray Cheng I fixed my post. See now. – Michael Rozenberg Jun 17 '17 at 5:42
• @MichaelRozenberg: Your answer is really unique, and correct. – DeepSea Jun 17 '17 at 6:28