I want to give this problem to high schooler students and I want to check if it is suitable for high school level.
Let $A$, $B$, $C$ be the sides of a triangle and $R$ be the triangle's circumcircle's radius. Prove that $$A+B+C\leq 3\sqrt{3}\,R$$
I would also like to get some insights on how people tend to approach this kind of problem, Your solutions are appreciated.
On
This is a proof without using any trigonometry, it only need Cauchy Schwarz inequality and a little bit of vector algebra over the plane.
Instead of capital letters, I will use $a, b, c$ to denote the sides of the triangle. I will reserve $A,B,C$ for the vertices of triangle and choose a coordinate system where circumcenter is located at origin. i.e. $|A| = |B| = |C| = R$.
By Cauchy Schwartz, we have
$$a + b + c = a \cdot 1 + b \cdot 1 + c\cdot 1 \le \sqrt{a^2+b^2+c^2}\sqrt{1^2 + 1^2 + 1^2}$$ Since $$\begin{align}a^2 + b^2 + c^2 &= |B-C|^2 + |C-A|^2 + |A-B|^2\\ &= 2(|A|^2 + |B|^2 + |C^2|) - 2(B\cdot C + C\cdot A + A\cdot B)\\ &= 3(|A|^2 + |B|^2 + |C^2|) - |A+B+C|^2\\ &\le 9R^2\end{align}$$
We find $$a+ b + c \le \sqrt{9R^2}\cdot\sqrt{3} = 3\sqrt{3}R$$
On
In the standard notation we need to prove that: $$a+b+c\leq3\sqrt3\cdot\frac{abc}{4S}$$ or $$27a^2b^2c^2\geq(a+b+c)^3\prod_{cyc}(a+b-c).$$ Now, let $a=y+z,$ $b=x+z$ and $c=x+y.$
Thus, $x$, $y$ and $z$ are positives and we need to prove that $$27\prod_{cyc}(x+y)\geq64(x+y+z)^3xyz.$$ But $$(x+y)(x+z)(y+z)\geq\frac{8}{9}(x+y+z)(xy+xz+yz)$$ it's $$\sum_{cyc}z(x-y)^2\geq0.$$ Id est, it's enough to prove that $$(xy+xz+yz)^2\geq3(x+y+z)xyz,$$ which is $$\sum_{cyc}z^2(x-y)^2\geq0.$$
Well, I'd start by rewriting the inequality by the sine rule (here $\alpha$ is the angle opposite of side $A$, and so on):
$$ \sin \alpha + \sin \beta + \sin \gamma \le \frac{3\sqrt{3}}{2},$$
so that is what we now want to prove; by Jensen's inequality (since $\sin x$ is concave in $[0, \pi]$), we have that
$$\sin \alpha + \sin \beta + \sin \gamma \le 3 \sin \frac{\alpha + \beta + \gamma}{3} = \frac{3\sqrt{3}}{2}.$$