Prove $ P ≤ \frac{\sqrt 3}{4}(abc)^\frac{2}{3} $ for triangle with sides $ a,b,c $ and area P

Question

Prove that for any triangle with sides $ a,b,c $ and area $P$ the following inequality holds:$ P ≤ \frac{√3}{4}(abc)^\frac{2}{3}$. Find all triangles for which equality holds.

What I've tried :

We've $ P ≤ \frac{√3}{4}(abc)^\frac{2}{3}$

=> $ 16P^2 ≤ 3(abc)^\frac{4}{3} $

Using heron's formula :

$$ (a+b+c)(b+c-a)(a+c-b)(a+b-c) ≤ 3(abc)^\frac{4}{3} $$ Further using A.M. - G.M. :

$$ (\frac{a+b+c}{2})^4 ≥ (a+b+c)(b+c-a)(a+c-b)(a+b-c) $$ But from here I'm not getting any idea how to show $ (\frac{a+b+c}{2})^4 ≤ 3(abc)^\frac{4}{3} $ .

Answer 1

I don't think you'll be able to use AM-GM as you tried as the inequality is pointing in the wrong direction. Instead, we'll make use of the following inequality for a triangle with vertices $A,B$ and $C$:

$$\sin A+\sin B+\sin C \leq \frac{3\sqrt{3}}{2}$$

This is usually proven with Jensen's inequality, but since you tagged this question as 'pre-calculus', here's a proof that just uses the sum-to-product identity. Note that if $\alpha$ and $\beta$ are angles in a triangle, then $\alpha + \beta < \pi$, and so

$$\sin \alpha + \sin \beta = 2 \sin \left( \frac{\alpha + \beta}{2} \right) \cos \left( \frac{\alpha - \beta}{2} \right) \le 2 \sin \left( \frac{\alpha + \beta}{2} \right)$$

Now suppose WLOG that $C$ is the smallest angle in the triangle, so that $C \le \frac{\pi}{3}$. Then we get that

$$\begin{align} \sin A + \sin B + \sin C + \sin \frac{\pi}{3} &\le 2 \sin \left( \frac{A+B}{2} \right) + 2 \sin \left( \frac{C+\frac{\pi}{3}}{2} \right) \\[2mm] &\le 4 \sin \left( \frac{A+B+C+\frac{\pi}{3}}{4} \right) \\[2mm] &= 4 \sin \left( \frac{\frac{4\pi}{3}}{4} \right) \\[2mm] &= 4 \sin \left( \frac{\pi}{3} \right) \end{align}$$

which then yields

$$\sin A + \sin B + \sin C \le 3 \sin \left( \frac{\pi}{3} \right) = \frac{3\sqrt{3}}{2}$$

Now to get the desired inequality, we use the area formula

$$P = \frac{1}{2} ab \sin C = \frac{1}{2} ac \sin B = \frac{1}{2} bc \sin A$$

Multiplying these three and then applying AM-GM leads to

$$\begin{align} P^3 &= \frac{1}{8} a^2b^2c^2 \sin A \sin B \sin C \\[2mm] &\le \frac{1}{8} a^2b^2c^2 \left( \frac{\sin A + \sin B + \sin C}{3} \right)^3 \\[2mm] &= \frac{1}{8} a^2b^2c^2 \left( \frac{\sqrt{3}}{2} \right)^3 \\[2mm] &= \frac{3 \sqrt{3}}{64} a^2b^2c^2 \end{align}$$

and so

$$P \le \frac{\sqrt{3}}{4} (abc)^{\frac{2}{3}}$$

To investigate equality, note that we made use of two inequalities in the proof. The first will become an equality if $\cos \left( \frac{\alpha - \beta}{2} \right)$ is equal to $1$, i.e. if $A=B$ and $C=\frac{\pi}{3}$, so all three angles would have to be equal; and I'm sure you know AM-GM will only be an equality if all terms are equal. Thus equality for our case will only hold if the triangles are equilateral.

Answer 2

Here is a proof using pqr method:

Let $x = (a + b - c)/2 > 0$ and $y = (b + c - a)/2 > 0$ and $z = (c + a - b)/2 > 0$. Then $a = z + x, b = x + y, c = y + z$.

By Heron's formula, we need to prove that $$(a + b + c)(a + b - c)(b + c - a)(c + a - b) \le 3(abc)^{4/3}$$ which is written as $$16(x + y + z) xyz \le 3[(x + y)(y + z)(z + x)]^{4/3}. \tag{1}$$

Let $p = x + y + z, q = xy + yz + zx, r = xyz$.

(1) is written as $$16p r \le 3(pq - r)^{4/3}. \tag{2}$$

Using $q^2 \ge 3pr$ and $p^2 \ge 3q$, we have $r \le \frac{q^2}{3p}$ and $$pq - r \ge pq - \frac{q^2}{3p} = \frac{q(3p^2 - q)}{3p} > 0.$$

It suffices to prove that $$16p \cdot \frac{q^2}{3p} \le 3\left(pq - \frac{q^2}{3p}\right)^{4/3}$$ or $$\frac{16}{3}q^2 \le 3\left(pq - \frac{q^2}{3p}\right)^{4/3}.$$

Using $p^2 \ge 3q$, we have $p \ge \sqrt{3q}$ and $$pq - \frac{q^2}{3p} \ge q\sqrt{3q} - \frac{q^2}{3\sqrt{3q}} = \frac89 q\sqrt{3q} > 0.$$ We have $$3\left(pq - \frac{q^2}{3p}\right)^{4/3} \ge 3\left(\frac89 q\sqrt{3q}\right)^{4/3} = \frac{16}{3}q^2.$$

We are done.