<s>[INST] Geometric inequality: $2r^2+8Rr \leq \frac{a^2+b^2+c^2}{2}$ Suppose $a$, $b$, and $c$ are the lengths of the sides of a triangle, and $R$ and $r$ are its circumradius and inradius respectively. How can one prove the following inequality? $$2r^2+8Rr \leq \frac{a^2+b^2+c^2}{2}$$
 [/INST] We need to prove that
$$2r^2+8Rr \leq \frac{a^2+b^2+c^2}{2}$$ or
$$2\cdot\frac{4S^2}{(a+b+c)^2}+8\cdot\frac{abc}{4S}\cdot\frac{2S}{a+b+c}\leq\frac{a^2+b^2+c^2}{2}$$ or
$$\frac{\prod\limits_{cyc}(a+b-c)}{2(a+b+c)}+\frac{4abc}{a+b+c}\leq\frac{a^2+b^2+c^2}{2}$$ or
$$(a^2+b^2+c^2)(a+b+c)-8abc+\sum_{cyc}\left(a^3-a^2b-a^2c+\frac{2}{3}abc\right)\geq0$$ or
$$a^3+b^3+c^3-3abc\geq0,$$
which is AM-GM.
Done!
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