Hadwiger-Finsler Inequality gives $a^2+b^2+c^2 \ge 4\sqrt{3}S$, which is stronger.

Easy. $\frac{a^2+b^2+c^2}{6}\geq\frac{ab+ac+bc}{6}$ and since $S=\frac{1}{2}absin\alpha$ and $sin\alpha\leq 1$ then $ab\geq 2S$. Similarly $bc\geq 2S$ and $ac\geq 2S$. Adding up we get $\frac{ab+ac+bc}{6}\geq\frac{2S+2S+2S}{6}=S$. Q.E.D.

Substitute S with expression involving a, b, c and it's right there.

But $equality$ doesn't follow.

heronbash it

I think that there is no equality case

Let $a\le b\le c$. $$a^2+b^2\ge2ab=4\cdot\frac{1}{2}ab\ge4\cdot\frac{1}{2}ab\cdot\sin C=4S$$$$\frac{a^2+b^2}{4}\ge S$$$$\frac{a^2+b^2+c^2}{6}\ge\frac{a^2+b^2}{4}\ge S$$However there cannot be equality, as if $\sin C=1$, then we cannot have $a=b=c$

@below, just realised the mistake.

WizardMath wrote: Kaskade wrote: $$\frac{a^2+b^2+c^2}{6}\ge\frac{a^2+b^2}{4}\ge S$$ I would say that this holds only for obtuse triangles. We assumed $a\le b\le c$

Sol:- Let $R$ be the circumcircumradius and $A,B,C$ be the angles opposite to sides a,b,c of the triangle. It is known that $S=abc/4R$ and $a=2R sin A, b=2R sin B , c=2R sin C$. Note that $S=\frac{abc}{4R} \leq \frac{a^2+b^2+c^2}{6} \iff sin A sin B sin C \leq \frac{sin^2 A + sin ^2 B + sin^2 C}{3}$. Note that by AM-GM $sin A sin B sin C \leq (sin A sin B sin C)^{\frac{2}{3}} \leq \frac{sin^2 A + sin ^2 B + sin^2 C}{3}$. So we are done

i want to envolve problem like this cause there arent any equality case and i will solve it with ravi.... $$\frac{a^2+b^2+c^2}{4\sqrt{3}} \geq S$$it follows Ravi $$x^2+y^2+z^2+xy+yz+zx \geq 2\sqrt{3(x+y+z)xyz}$$$$x^2+y^2+z^2 \geq xy+yz+zx$$$$xy+yz+zx \geq 2\sqrt{3(x+y+z)xyz}$$and we are done

By Hadwiger-Finsler, $\frac{2\sqrt{3}}{6}\le\frac{a^2+b^2+c^2}{6}$, since $\frac{2\sqrt{3}}{6} \ge 1$, hence, proved. if you click on the period, you can see proof of hadwiger-finsler

$$2S \leq absin\alpha \leq ab (1)$$$$2S \leq bcsin\beta \leq bc (2)$$$$2S \leq casin\gamma \leq ca (3)$$ (1)+(2)+(3) gives us $$6S \leq ab+bc+ca \leq a^2+b^2+c^2$$ Divide both sides by 6 and we're done

Sadigly wrote: $$2S \leq absin\alpha \leq ab (1)$$$$2S \leq bcsin\beta \leq bc (2)$$$$2S \leq casin\gamma \leq ca (3)$$ (1)+(2)+(3) gives us $$6S \leq ab+bc+ca \leq a^2+b^2+c^2$$ Divide both sides by 6 and we're done When does the equality occur? $\alpha=\beta=\gamma=90\circ$?

m4thbl3nd3r wrote: Sadigly wrote: $$2S \leq absin\alpha \leq ab (1)$$$$2S \leq bcsin\beta \leq bc (2)$$$$2S \leq casin\gamma \leq ca (3)$$ (1)+(2)+(3) gives us $$6S \leq ab+bc+ca \leq a^2+b^2+c^2$$ Divide both sides by 6 and we're done When does the equality occur? $\alpha=\beta=\gamma=90^\circ$?

$6\to 4\sqrt{3}$ is a stronger inequality.