Triangle inequality involving side lengths and circumradius

Question

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

Answer 1

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?

Hint: This can be easily proved using Jensen's inequality.

Answer 2

We know that $R \ge 2r$$R \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}$