Alternate proof for $a^2+b^2+c^2\le 9R^2$

Question

As I studying geometric inequalities, one of those famous inequalities is $$a^2+b^2+c^2\le 9R^2$$

I did some research and I found that there is a proof (not exactly the this inequality but an useful identity) of this on geometry revisited book section 1.7. the identity is $$OH^2=9R^2-(a^2+b^2+c^2)$$

where $H$ is orthocenter and $O$ is circumcenter. the proof of this identity uses Stewart's theorem, Euler line and ... . I find the proof not very nice and a little bit brute force. I want to know is there any different proof for it? and what is the name of this inequality?

Answer 1

Sincne $\sin^2A+\sin^2B+\sin^2C=2+2\cos A\cos B\cos C$ and $a=2R\sin A ,b=2R\sin B, c=2R\sin C$ in a triangle: $$a^2+b^2+c^2=4R^2(\sin^2A+\sin^2B+\sin^2C)=4R^2(2+2\cos A\cos B\cos C)$$ Since, $\cos A\cos B\cos C\leq \frac18$ or, $\sin^2A+\sin^2B+\sin^2C\leq\frac94$ both of which can be proved independently. $$a^2+b^2+c^2\leq9R^2$$

Answer 2

$\def\vec#1{\overrightarrow{#1}}$ From $\triangle OBC$, by squaring $\vec{BC}=\vec{OC}-\vec{OB}$ we get $a^2=2R^2-2\vec{OB}\cdot\vec{OC}$, or $$ 2\vec{OB}\cdot\vec{OC}=2R^2-a^2 $$ and similarly $$ 2\vec{OC}\cdot\vec{OA}=2R^2-b^2,\quad 2\vec{OA}\cdot\vec{OB}=2R^2-c^2. $$ It follows that $$ \left\Vert\vec{OA}+\vec{OB}+\vec{OC}\right\Vert^2=9R^2-(a^2+b^2+c^2) $$ Which is clearly positive.

Remark. To prove the inequality, we do not need the fact that $\vec{OH}=\vec{OA}+\vec{OB}+\vec{OC}$.

Answer 3

Let $a^2=x$, $b^2=y$ and $c^2=z$.

Hence, we need to prove that $$a^2+b^2+c^2\leq\frac{9a^2b^2c^2}{16S^2}$$ or $$a^2+b^2+c^2\leq\frac{9a^2b^2c^2}{\sum\limits_{cyc}(2a^2b^2-a^4)}$$ or $$9xyz\geq\sum_{cyc}x\sum_{cyc}(2xy-x^2)$$ or $$\sum_{cyc}(x^3-x^2y-x^2z+xyz)\geq0,$$ which is Schur.

Done!