Properties of Triangle

\frac{s^{2}}{4}=\frac{(a+b+c)^{}2}{16}=\frac{a^{2}+b^{2}+c^{2}+2ab+2bc+2ca}{16}}{}

a^{2}+b^{2}+c^{2} \geq ab + bc + ca\text{ (Rearrangement inequality)}

\implies \frac{s^{2}}{4} \geq \frac{3(ab + bc+ca)}{16}=\frac{3\cdot 2}{16}\Delta\left(\frac{1}{sinC}+\frac{1}{sinA}+\frac{1}{sinB}\right)
(\because \frac{1}{2}a\cdot bsinC =\Delta,\frac{1}{2}b\cdot csinA =\Delta,\frac{1}{2}c\cdot asinB =\Delta)

\text{Being the angle's of a triangle A+B+C=180}

\text{Since }sin\theta \text{ is a convex function (i.e. on differentiating it} \\\text{twice we get a a function which is}>0\text{ for all values of }\theta \text{ b/w 0 to 180)}

\text{We can apply the Jensen's inequality,}

f(a)+f(b)+f(c)\geq 3f\left(\frac{a+b+c}{3}\right)=3\frac{1}{sin(60)}=2\sqrt{3}

\therefore \frac{s^{2}}{4} \geq \frac{3\times 2}{16}\Delta 2\sqrt{3}=\frac{3\sqrt{3}\Delta}{4} > \Delta

Small mistake. sinx is concave, and i guess f(x) = cosecx here?

@Sourish ...Well I intended to write 1sinx=cosec x.
cosex is convex ,sinx is concave.
Anyways the rest is the same.
Thanks 4 pointing it out.

@Sir: I think that s²4 should be strictly greater than Δ since the 4 terms can never be equal.

This is because s can never be equal to s-a.

s²4 is minimum when the triangle is equilateral in which case it is:3√3Δ4...:)