crazyfehmy wrote: Let $a, b, c$ be the side-lengths of a triangle, $r$ be the inradius and $r_a, r_b, r_c$ be the corresponding exradius. Show that \[ \frac{a+b+c}{\sqrt{a^2+b^2+c^2}} \leq 2 \cdot \frac{\sqrt{{r_a}^2+{r_b}^2+{r_c}^2}}{r_a+r_b+r_c-3r} \] $a+b+c=2p$ $p-a=x$,$p-b=y$,$p-c=z$ $\Longrightarrow p=x+y+z,x,y,z >0,a=y+z,b=x+z,c=x+y$ we have $\sqrt{p(p-a)(p-b)(p-c)}=S_{\triangle ABC}=pr=(p-a)r_{a}=(p-b)r_{b}=(p-c)r_{c} $ so $r=\sqrt\frac{xyz}{x+y+z},r_{a}=\sqrt\frac{yz(x+y+z)}{x},r_{b}=\sqrt\frac{xz(x+y+z)}{y},r_{c}=\sqrt\frac{yx(x+y+z)}{z}$ so we need prove $2\frac{\sum x}{\sqrt{\sum{(x+y)^{2}}}} \le 2\frac{\sqrt{\sum\frac{(x+y+z)xy}{z}}}{\sum\sqrt\frac{(x+y+z)xy}{z}-3\sqrt\frac{xyz}{\sum x}} $ $\iff \frac{\sum x}{\sqrt{\sum{(x+y)^{2}}}} \le \frac{\frac{\sqrt{\sum x \sum{x^{2}y^{2}}}}{\sqrt{xyz}}}{\frac{\sum x \sum{xy}-3xyz}{\sqrt{(x+y+z)xyz}}}$ $\iff \sqrt{\sum {(x+y)^2}\sum {x^2y^2}} \ge \sum_{sym}{x^2y} =\sum{xy(x+y)}$ it is true by use cauchy inequality

Nice proof, caoquyetthang. My proof is very similar. Using the Cauchy-Schwarz inequality, we need to show that \[2(ar_a+br_b+cr_c)\geq (a+b+c)(r_a+r_b+r_c-3r).\] Since $r_a=\frac{\Delta}{s-a};$ the last inequality can be rephrased as \[2\sum_{cyc}\frac{a}{s-a}\geq (a+b+c)\left(\sum_{cyc}\frac 1{s-a}-\frac{3}{s}\right);\] Or, \[2\sum_{cyc}\frac{a}{s-a}\geq \sum_{cyc}\frac{a}{s-a}+\sum_{cyc}\frac{b+c}{s-a}-\frac{3(a+b+c)}{s};\] Or, \[\sum_{cyc}\frac{a}{s-a}\geq \sum_{cyc}\frac{2s-a}{s-a}-6;\] Or, \[\sum_{cyc}\frac{a}{s-a}-\sum_{cyc}\frac{s}{s-a}\geq -3;\] Which is actually an identity. Equality holds if and only if $\frac{r_a}{a}=\frac{r_b}{b}=\frac{r_c}{c}\iff a=b=c.\Box$

By Cauchy-Schwarz,we just need to show that\[\begin{aglined}2\sum ar_a+3r\sum a=\sum a\sum r_a\]\[\iff 3r\sum a=(b+c-a)r_a+(c+a-b)r_b+(a+b-c)r_c\]Notice that\[2S_{\triangle ABC}=r\sum a=(b+c-a)r_a=(c+a-b)r_b=(a+b-c)r_c\]So it's obviously true.

Nirvanacs wrote: By Cauchy-Schwarz,we just need to show that\[\begin{aglined}2\sum ar_a+3r\sum a=\sum a\sum r_a\]\[\iff 3r\sum a=(b+c-a)r_a+(c+a-b)r_b+(a+b-c)r_c\]Notice that\[2S_{\triangle ABC}=r\sum a=(b+c-a)r_a=(c+a-b)r_b=(a+b-c)r_c\]So it's obviously true. beautiful！

MY SOLUTION= first we substitute $r_a=\frac{\triangle}{s-a}$ etc. and $r=\frac{\triangle}{s}$ where $\triangle$ represent area of triangle $ABC$. so substituting we have $ 2 \cdot \frac{\sqrt{{r_a}^2+{r_b}^2+{r_c}^2}}{r_a+r_b+r_c-3r} = \frac{s(\sqrt {\sum (s-a)^2(s-b)^2})}{\sum s(s-a)(s-b)-3(s-a)(s-b)(s-c)}$ also as $2s=a+b+c$ our inequality transforms into $\frac{1}{\sqrt{a^2+b^2+c^2}}\leq \frac{\sqrt {\sum (s-a)^2(s-b)^2}}{\sum s(s-a)(s-b)-3(s-a)(s-b)(s-c)}$ but by cauchy schwarz we have $(\sqrt{a^2+b^2+c^2})(\sqrt {\sum(s-a)^2(s-b)^2})\ge a(s-b)(s-c)+b(s-a)(s-c)+c(s-a)(s-b)=\sum s(s-a)(s-b)-3(s-a)(s-b)(s-c)$ hence we are done