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Give a triangle with the perimeter $2$. Show that: $$\max\left \{ 1,\,3\sqrt[3]{r} \right \}\leq m_{a}+ m_{b}+ m_{c}<\frac{3}{\sqrt{2}}$$ My try: We have: $$m_{a},\,m_{b},\,m_{c}= \frac{1}{2}\sqrt{2\,b^{2}+ 2\,c^{2}- a^{2}},\,\frac{1}{2}\sqrt{2\,a^{2}+ 2\,b^{2}- b^{2}},\,\frac{1}{2}\sqrt{2\,c^{2}+ 2\,a^{2}- c^{2}}$$ & $$\left ( a+ b+ c \right )^{3}\geq 3\sum\limits_{cycl}a\left ( 2\,b^{2}+ 2\,c^{2}- a^{2} \right )$$ $$\Leftarrow \sum\limits_{cycl}\left ( 4\,a^3-3\,a^2b-3\,a^2c+2\,abc \right )\geq 0$$ $$\Leftarrow 2\sum\limits_{cycl}\left ( a^3-a^2b-a^2c+abc \right )+\sum\limits_{cycl}\left ( 2\,a^3-a^2b-a^2c \right )\geq 0$$ which is Schur and the last is Muirhead. I can't continue with my opinion! I need to the help! Thanks!

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In the standard notation by the triangle inequality we obtain: $$\sum_{cyc}m_a<\sum_{cyc}\frac{b+c}{2}=a+b+c=2<\frac{3}{\sqrt2}.$$ The left inequality.

Let $a=y+z$, $b=x+z$ and $c=x+y$.

Hence, $x$, $y$ and $z$ are positives and $$\sum_{cyc}m_a=\frac{1}{2}\sqrt{2b^2+2c^2-a^2}=\frac{1}{2}\sum_{cyc}\sqrt{2(x+z)^2+2(x+y)^2-(y+z)^2}=$$ $$=\frac{1}{2}\sum_{cyc}\sqrt{4x(x+y+z)+(y-z)^2}\geq\frac{1}{2}\sum_{cyc}\sqrt{4x(x+y+z)}=$$ $$=(\sqrt{x}+\sqrt{y}+\sqrt{z})\sqrt{x+y+z}.$$ We see that $$(\sqrt{x}+\sqrt{y}+\sqrt{z})\sqrt{x+y+z}\geq1$$ it's $$(\sqrt{x}+\sqrt{y}+\sqrt{z})\sqrt{x+y+z}\geq x+y+z$$ or $$\sqrt{x}+\sqrt{y}+\sqrt{z}\geq\sqrt{x+y+z},$$ which is obvious.

Id est, it's enough to prove that $$(\sqrt{x}+\sqrt{y}+\sqrt{z})\sqrt{x+y+z}\geq3\sqrt[3]r$$ or $$(\sqrt{x}+\sqrt{y}+\sqrt{z})\sqrt{x+y+z}\geq3\sqrt[3]{\frac{2S}{a+b+c}\cdot\frac{(a+b+c)^2}{4}}$$ or $$(\sqrt{x}+\sqrt{y}+\sqrt{z})\sqrt{x+y+z}\geq3\sqrt[3]{(x+y+z)\sqrt{xyz(x+y+z)}}$$ or $$\sqrt{x}+\sqrt{y}+\sqrt{z}\geq3\sqrt[6]{xyz},$$ which is AM-GM.

Done!

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  • $\begingroup$ @Cô_Ngốc_Lớp_Trưởng I solved this problem for you. Why you down-voted? $\endgroup$ – Michael Rozenberg Feb 17 at 6:10
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    $\begingroup$ So you can see ... $\endgroup$ – user685500 May 11 at 10:32

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