Triangle inequality $\frac{ab}{a^{2}+ b^{2}}+ \frac{bc}{b^{2}+ c^{2}}+ \frac{ca}{c^{2}+ a^{2}}\geq \frac{1}{2}+ \frac{2r}{R}$

Question

Suppose that $a, b, c$ be the lengths of 3 sides of a triangle, $r$ is the inradius of the circle and $R$ is the circumradius of the circle. Prove that $$\frac{ab}{a^{2}+ b^{2}}+ \frac{bc}{b^{2}+ c^{2}}+ \frac{ca}{c^{2}+ a^{2}}\geq \frac{1}{2}+ \frac{2r}{R}$$ It' s so amazing, I tried to use Euler and Gerretsen's inequality.

Source

The following AoPS post qualifies as a source. Please see the discussion following the question.

Results which can be used

Euler's inequality : $R \geq 2r$, where $R,r$ are the circumradius and inradius respectively.

Gerritsen's inequality : Let $R,r$ be as above and $s$ be the semiperimeter of the triangle. Then, $$ 16Rr - 5r^2 \leq s^2 \leq 4R^2+4Rr+3r^2 $$

See here for a proof of Gerritsen's inequality.

Answer 1

In the standard notation we need to prove that: $$\sum_{cyc}\frac{ab}{a^2+b^2}\geq\frac{1}{2}+\frac{\frac{4S}{a+b+c}}{\frac{abc}{4S}}$$ or $$\sum_{cyc}\left(\frac{ab}{a^2+b^2}-\frac{1}{2}\right)\geq\frac{16S^2}{abc(a+b+c)}-1$$ or $$1-\frac{\prod\limits_{cyc}(a+b-c)}{abc}\geq\sum_{cyc}\frac{(a-b)^2}{2(a^2+b^2)}$$ or $$\frac{2\sum\limits_{cyc}(a^3-a^2b-a^2c+abc)}{abc}\geq\sum_{cyc}\frac{(a-b)^2}{a^2+b^2}$$ or $$\frac{\sum\limits_{cyc}(2a^3-2abc-2(a^2c+b^2c-2abc))}{abc}\geq\sum_{cyc}\frac{(a-b)^2}{a^2+b^2}$$ or $$\frac{\sum\limits_{cyc}(a-b)^2(a+b-c)}{abc}\geq\sum_{cyc}\frac{(a-b)^2}{a^2+b^2}$$ or $$\sum_{cyc}(a-b)^2\left(a+b-c-\frac{abc}{a^2+b^2}\right)\geq0.$$ Now, let $a\geq b\geq c$.

Thus, by AM-GM $$a+b-c-\frac{abc}{a^2+b^2}\geq a+b-c-\frac{abc}{2ab}=a+b-1.5c>0$$ and $$a+c-b-\frac{abc}{a^2+c^2}\geq c-\frac{abc}{a^2+c^2}=c\left(1-\frac{ab}{a^2+c^2}\right)>0.$$ Id est, since $c>a-b$, we obtain: $$\sum_{cyc}(a-b)^2\left(a+b-c-\frac{abc}{a^2+b^2}\right)\geq$$ $$\geq(a-c)^2\left(a+c-b-\frac{abc}{a^2+c^2}\right)+(b-c)^2\left(b+c-a-\frac{abc}{b^2+c^2}\right)\geq$$ $$\geq(b-c)^2\left(a+c-b-\frac{abc}{a^2+c^2}\right)+(b-c)^2\left(b+c-a-\frac{abc}{b^2+c^2}\right)=$$ $$=c(b-c)^2\left(2-ab\left(\frac{1}{a^2+c^2}+\frac{1}{b^2+c^2}\right)\right)>$$ $$>c(b-c)^2\left(2-ab\left(\frac{1}{a^2+(a-b)^2}+\frac{1}{b^2+(a-b)^2}\right)\right)=$$ $$=\frac{c(b-c)^2(a-b)^2(4a^2-7ab+4b^2)}{(a^2+(a-b)^2)(b^2+(a-b)^2)}\geq0.$$

Answer 2

I can suggest a weaker inequality $$\frac{c^{2}}{a^{2}+ b^{2}}+ \frac{a^{2}}{b^{2}+ c^{2}}+ \frac{b^{2}}{c^{2}+ a^{2}}\leq \frac{2R- r}{2R}$$ Proof $$\begin{align*} \frac{a^{2}}{b^{2}+ c^{2}}+ \frac{b^{2}}{c^{2}+ a^{2}}+ \frac{c^{2}}{a^{2}+ b^{2}}\leq \frac{a^{3}+ b^{3}+ c^{3}}{2abc}\\ a^{3}+ b^{3}+ c^{3}= 2s^{3}- 12R- 6r\\ abc= 4R\\ \Leftrightarrow LHS\leq \frac{s^{2}- 6Rr- 3r^{2}}{4Rr}\leq_{GERRETSEN} \frac{\left ( 4R^{2}+ 4Rr+ 3r^{2} \right )- 6Rr- 3r^{2}}{4Rr}= \frac{2R- r}{2r}\\ \end{align*}$$

$$\begin{align*} \frac{a^{2}}{b^{2}+ c^{2}}+ \frac{b^{2}}{c^{2}+ a^{2}}+ \frac{c^{2}}{a^{2}+ b^{2}}\geq \frac{\left ( b- c \right )^{2}}{b^{2}+ c^{2}}+ \frac{\left ( c- a \right )^{2}}{c^{2}+ a^{2}}+ \frac{\left ( a- b \right )^{2}}{a^{2}+ b^{2}}\\ \Leftrightarrow \frac{bc}{b^{2}+ c^{2}}+ \frac{ca}{c^{2}+ a^{2}}+ \frac{ab}{a^{2}+b^{2}}\geq \frac{7r- 2R}{4r}\ngeqslant \frac{1}{2}+ \frac{2r}{R} \end{align*}$$ I can't succeed in proving this inequality by use it! How about another solution? I want to see more.

Answer 3

$$\frac{c^{2}}{a^{2}+ b^{2}}+ \frac{a^{2}}{b^{2}+ c^{2}}+ \frac{b^{2}}{c^{2}+ a^{2}}\leq \frac{2R- r}{2R}$$ Proof