# Triangle inequality involving side lengths and circumradius

For a $$\triangle ABC$$ with side lengths $$a,b,c$$ and circumradius $$R$$ prove that-

$$a+b+c\leq 3\sqrt3 R$$

Now we know that $$R=\frac{abc}{4\Delta}$$ where $$\Delta$$ denotes area of triangle. So I tried to reduce inequality as follows to find hint of using $$AM-GM$$

$$a+b+c\leq 3\sqrt3 \frac{abc}{4\Delta}$$

$$(a+b+c)^2\leq 27 \frac{(abc)^2}{16\Delta^2}$$

$$\displaystyle(a+b+c)^3(a+b-c)(b+c-a)(c+a-b)\leq \frac{27(abc)^2}{16}$$

Now I know that which terms I have to incorporate in $$AM-GM$$ but I'm not able to correctly use them

Please provide me hint so that I can proceed forward

• Several hi(n)ts from approach0.xyz, as expected.
– dxiv
Jun 7 at 6:37

Use the sine rule $$a = 2R\sin A$$ $$b = 2R\sin B$$ $$c = 2R\sin C$$ Now, you have to prove that $$2R\sin A + 2R\sin B + 2R\sin C \le 3\sqrt3R$$ $$\sin A + \sin B+ \sin C \le \dfrac{3\sqrt3}{2}$$ Can you complete it from here?
• Yes now I can take it forward since$\ sin x$ is convex up for the interval $(0,\pi)$, Therefore $(\sin A+\sin B+\sin c)/3 \leq \sin( (A+B+C)/3)=\sqrt{3}/2$ Jun 7 at 6:39
We know that $$R \ge 2rR \ge 2r=\frac{2Δ}{s}=2\sqrt{\frac{(s-a)(s-b)(s-c)}{s}}\ge 2(\frac{3s-a-b-c}{3})^\frac32\cdot \frac{1}{\sqrt{s}}=2(\frac{s}{3})^\frac32\cdot \frac{1}{\sqrt{s}}=\frac{2s}{3\sqrt3}=\frac{a+b+c}{3\sqrt3}$$
• @LalitTolani Here's proof why $R\ge 2r$ (math.stackexchange.com/a/171641/742113) Jun 7 at 6:49