$2\sqrt3 h\geq$ perimeter Problem:
Let $ABC$ be an acute triangle and $h$ be the longest height. Show that $2\sqrt3h\geq$ the perimeter of $ABC$.

I have a solution using geometric arguments, where I reduce the problem into isoceles triangle and bash out the length.

Here is my problem: When I tried to use algebraic approaches, I stuck at the following:
$$3(b+c-a)(c+a-b)(a+b-c)\geq c^2(a+b+c)$$
where $a,b,c$ is the side length with $c$ the minimum.
I wonder if this is doable, and if so, how? Thanks a lot.
 A: Note that you haven't used the fact that $ABC$ is an acute triangle. In particular, the inequality is no longer true for some (very) obtuse triangles (EG $c = 1, b = 1, a = 1.5$).

*

*Suppose $c = 1$ is fixed.

*Suppose $ a + b = k \geq 2 $ is fixed. Notice that the RHS is fixed $=k+1$. How can we minimize the LHS?

*

*WLOG $ a \geq b \geq c = 1$

*$(a+b-c) = k-1$ is fixed.

*$(b+c -a )(c+a-b) = 1^2 - (a-b)^2$

*Note that the triangle inequality gives us $a-b < c = 1$, so this value is positive.

*So LHS$=3 ( 1 - (a-b)^2)(k-1)$. To minimize this, we have to maximize $a-b$.



*Case 1: $ 2 \leq k < 2.4$

*

*Since $ b \geq 1$,  $ a -b  = k - 2 b \leq k - 2$

*It remains to verify that $ 3 ( 1 - (k-2)^2)(k-1) \geq (k+1)$.

*This is true for the domain, EG Wolfram.

*Equality holds iff $k = 2$, which is the equilateral triangle case.



*Case 2: $ 2.4 \leq k $.

*

*Acute triangle condition: $a^2 \leq b^2 + c^2$

*Since $a^2 - b^2 \leq 1$ and $ a +b = k$ so $a-b \leq \frac{1}{k}$.

*It remains to verify that $3 ( 1 - \frac{1}{k^2} ) (k-1) \geq (k+1)$.

*Again, this is true for the domain, EG Wolfram

*Equality doesn't hold here.



Notes

*

*Yes, the cases were chosen to fit the inequality domains. We could have used a slightly different split since the domains overlapped.

*I would love to see a solution that doesn't break down into cases, perhaps by showing that $ (a-b)^2 \leq \frac{2}{3} \frac{k-2}{k-1}$.

A: Geometric solution. As shown below, drawn is the longest height.

The longest height should be on the shortest side, which is opposite to the smallest angle, thus $\alpha+\beta\le\frac\pi2-\alpha$, $\frac\pi2-\beta$, or $2\alpha+\beta$, $\alpha+2\beta\le\frac\pi2$. If we increase $\alpha$ until $2\alpha+\beta$ or $\alpha+2\beta$ reaches $\frac\pi2$, the perimeter increases while the height remains unchanged. It suffices to prove for such cases.

In fact, now the triangle is isosceles, because $2\alpha+\beta=\frac\pi2$ (resp. $\alpha+2\beta=\frac\pi2$) implies $\alpha+\beta=\frac\pi2-\alpha$ (resp. $\alpha+\beta=\frac\pi2-\beta$). Let the base have length $m$, the legs length $n\leq m$, and the angle opposite to the base be $\theta\in\left[\frac\pi3,\frac\pi2\right)$. $h$ should be on the legs. Then
\begin{align*}
&\text{original problem}\iff2\sqrt3h\ge n+n+m\\[2pt]
\iff&2\sqrt3n\sin\theta\ge2n+m\iff2\sqrt3\sin\theta\ge2+\frac mn\\[2pt]
\iff&2\sqrt3\sin\theta\ge2+\frac{\sin\theta}{\vphantom\strut\sin\frac{\pi-\theta}2}\iff\sqrt3\sin\theta\ge1+\sin\frac\theta2,
\end{align*}
which is easily proved. We're done.
A: 
Assume that without loss of any generality $a\gt b\gt c$. Then $h_c$, the altitude-$C$, is the longest height. As shown in the diagram, the foot of this altitude divide the side $c$ into two parts. We have,
$$\cot\left(A\right)=\dfrac{x}{h_c} \qquad\text{and}\qquad \cot\left(B\right)=\dfrac{c-x}{h_c}.$$
Therefore, we shall write,
$$c=c+\left(c-x\right)=h_c\left(\cot\left(A\right)+ \cot\left(B\right)\right)$$
We can obtain similar expressions for $a$ and $b$ in terms of $h_a$ and $h_b$ respectively.
$$a= h_a\left(\cot\left(B\right)+ \cot\left(C\right)\right)$$
$$b= h_b\left(\cot\left(C\right)+ \cot\left(A\right)\right)$$
$$\therefore\quad  p=a+b+c= h_a\left(\cot\left(B\right)+ \cot\left(C\right)\right)+ h_b\left(\cot\left(C\right)+ \cot\left(A\right)\right)+ h_c\left(\cot\left(A\right)+ \cot\left(B\right)\right)$$
Since $h_a \lt h_b \lt h_c$, we can write,
$$p\le 2h_c\left(\cot\left(A\right)+ \cot\left(B\right)+ \cot\left(C\right)\right).$$
Please note that the equality holds only for the equilateral triangle.
According to Wolfram MathWorld's $6^{\text{th}}$ triangle inequality, $\space\cot\left(A\right)+ \cot\left(B\right)+ \cot\left(C\right)\ge\sqrt{3}$.
$$\therefore\quad  p\le 2\sqrt{3}h_c. $$
A: Assuming that the shortest side is $c$, and the perimeter is $P$, the locus of $C$ is an ellipse whose foci are $A$ and $B$. For maximum $h$, $C$ is obviously at the minor axis of the ellipse. That means that the triangle is isosceles ($a=b$).
Since the triangle is acute and $\hat C$ is the lesser angle, that means that $\hat A=\hat B\ge\pi/3$.
Also,
$$\frac ha=\frac hb=\sin\hat A\ge\frac{\sqrt 3}2\implies \frac ah=\frac bh\le\frac{2\sqrt 3}3$$
$$\frac hc=\frac12\frac h{c/2}=\frac12\tan\hat A\ge\frac{\sqrt3}2\implies\frac ch\le\frac{2\sqrt 3}3$$
So
$$\frac{a+b+c}h\le2\sqrt 3$$ QED
A: Not entirely complete but even I've tried and I am also stuck at an inequality, I took $a$ as the smallest side
$h = \frac{2\triangle}{a} \ \  \  and \ \ 
4R\triangle = abc\\
h = \frac{bc}{2R}\\
h = \frac{bc\ sina}{a}\\
$
According to the question
$$\ \ \ \ \ \ \ \ \ \  \ \ \ \ \ \ 2\sqrt3h > 2s \\\\
 \ \ \ \ \ \frac{\sqrt3bcsina}{a} > s \\ 
 \ \ \ \ \ \ \ \ \ \  \ \ \ \ \  \ \ \ \ \  \ sin(a) > \frac{as}{\sqrt3 bc}\\
 \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ sin^2(a) > \frac{a^2s^2}{3b^2c^2}\\
 \ \ \ \ \ \ \ \ \ \  \ \ \ \ \ 1 - cos^2(a) > \frac{a^2s^2}{3b^2c^2}\\
1 - \frac{(b^2 + c^2 - a^2)^2}{4b^2c^2} > \frac{a^2s^2}{3b^2c^2}\\ 
12b^2c^2 - 3(b^2 + c^2 - a^2)^2 > 4a^2s^2 \\
 \ \ \ \ \ \ \ \ \ \ 12b^2c^2 - 3(b^2 + c^2 - a^2)^2 > a^2(a+b+c)^2
$$
Simplyfing it is
$6\displaystyle \sum_{cyc}a^2 b^2 > 3(\displaystyle \sum_{cyc}a^4) + a^2(\displaystyle \sum_{cyc}a^2) + 2a^2\displaystyle \sum_{cyc}(ab)$
Objective is to prove $a<b$ and $a<c$ I suppose but I've had no luck so far, I'll comment if I've managed to find something.
