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? $$2r^2+8Rr \leq \frac{a^2+b^2+c^2}{2}$$

We know that 
$$\begin{align*}
a+b+c&=2s,\\
ab+bc+ca&=s^2+r^2+4rR,
\end{align*}$$
where $s$ is semiperimeter, $r$ is inradius and $R$ is circumradius; 
see e.g., Manfrino, Ortega, Delgado - Inequalities: A Mathematical Olympiad Approach, Section 2.5.
Using these inequalities we can rewrite the LHS as
$$
\begin{align*}
2(r^2+4rR) &= 2(ab+bc+ca-s^2) \\
 &= 2(ab+bc+ca)-\frac{(a+b+c)^2}2 \\
 &= ab+bc+ca-\frac{a^2+b^2+c^2}2.
\end{align*}$$
After this the inequality we want to prove becomes
$$ab+bc+ca\le a^2+b^2+c^2,$$
which follows easily from AM-GM inequality. (Simply add the inequalities $ab\le \frac{a^2+b^2}2$, $ac\le \frac{a^2+c^2}2$ and $bc\le \frac{b^2+c^2}2$. )  See also this question for more proofs of the last inequality.

For the sake of completeness, let me copy here the proofs of these equalities as they are given in the book I mentioned above. (I hope that such a short excerpt still qualifies as a fair use. I tried to find a proof of them online, but I did not succeed.)

$$
\begin{align*}
a + b + c &= 2s, \tag{2.5}\\
ab + bc + ca &= s^2 + r^2 + 4rR, \tag{2.6}\\
abc &= 4Rrs. \tag{2.7}
\end{align*}
$$
  The first is the definition of $s$ and the third follows from the fact that the area of
  the triangle is $\frac{abc}{4R}=rs$. Using Heron’s formula for the area of a triangle, we have
  the relationship $s(s - a)(s - b)(s - c) = r^2s^2$, hence
  $$s^3 - (a + b + c)s^2 + (ab + bc + ca)s - abc = r^2s.$$
  If we substitute $(2.5)$ and $(2.7)$ in this equality, after simplifying we get that
  $$ ab + bc + ca = s^2 + r^2 + 4Rr.$$ 

