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today i was studying geometric inequalities and I saw this inequality $$R \ge 2r$$

unfortunately the book did not provided any prove or further explanations. So I just did a little research about it. I find that the name of inequality is euler triangle inequality and there was a simple proof about it

If $O$ is circumcenter of $\triangle ABC$ and $I$ is incenter of it and $OI=d$ then $$d^2=R(R-2r)$$

$R$ is circumradius and $r$ is inradius. from this identity we can drive $R-2r\ge0 \Rightarrow R\ge2r$

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but I did not liked this proof. so i continued my research till i found this from MAA.

first the author proves 3 lemma and using them simply proves the inequality. my problem was on understanding second lemma. here is a description about it

The second,

whose proof uses a rectangle composed of triangles similar to the right triangles

in FIGURE 1(b), expresses the product $xyz$ in terms of the inradius $r$ and the sum

$x + y + z$.

Figure 1(b)

And the lemma proof

LEMMA 2. $xyz = r^2(x + y + z)$.

Proof. Letting $w$ denote $\sqrt{r^2+x^2}$,we have

I can’t understand what sides are given in figure 3 and I can’t reach $r^2(x+y)$ (i tried using Pythagorean but i was not succesful).

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The “$\alpha$” triangle on the left of Figure 3 is an “$\alpha$” triangle from Figure 1b, but scaled by $yz$; so, instead of side-lengths $r$, $x$, $w \;(=\sqrt{r^2+x^2})$, it has side-lengths $r\cdot yz$, $x \cdot yz$, $w \cdot yz$.

Likewise, the “$\alpha$” triangle in the upper-right of Figure 3 is scaled by $rz$ to have side-lengths $x\cdot rz$, $r \cdot rz$, $w\cdot rz$.

This gives the “$\beta$” triangle a scale factor of $wz$ (leg-lengths $y\cdot wz$ and $r\cdot wz$); and, from comparing the top of the rectangle to the bottom, we have that the “$\gamma$” triangle has scale factor $r(x+y)$ (leg-lengths $z\cdot r(x+y)$ and $r\cdot r(x+y)$).

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