For $\triangle ABC$ are $AB=c, BC=a, CA=b$ and $AM_1=m_a, AM_2=m_b, AM_3=m_c$ are medians. Find minimum and maximum of $$\frac {m_a}{b+c}+\frac {m_b}{c+a}+ \frac {m_c}{a+b} $$
My work so far:
$$\frac{b+c}2>m_a$$ Then
$$\frac {m_a}{b+c}+\frac {m_b}{c+a}+ \frac {m_c}{a+b} <\frac32$$
If $a=b=1$ and $c\rightarrow2^-$ then $m_a=m_b\rightarrow\frac{3}{2}$ and $m_c\rightarrow0^+$.
Thus, $\sum\limits_{cyc}\frac{m_a}{b+c}\rightarrow1.$
We'll prove that it's an infimum.
Indeed, let $a=y+z$, $b=x+z$ and $c=x+y$.
Hence, we need to prove that $$\sum_{cyc}\frac{\sqrt{2b^2+2c^2-a^2}}{b+c}\geq2$$ or $$\sum_{cyc}\frac{\sqrt{4x(x+y+z)+(y-z)^2}}{2x+y+z}\geq2$$ or $$\sum_{cyc}\left(\tfrac{4x(x+y+z)+(y-z)^2}{(2x+y+z)^2}+\tfrac{2\sqrt{(4x(x+y+z)+(y-z)^2)(4y(x+y+z)+(x-z)^2)}}{(2x+y+z)(2y+x+z)}\right)\geq4.$$ Now, by C-S and AM-GM $$\sqrt{(4x(x+y+z)+(y-z)^2)(4y(x+y+z)+(x-z)^2)}\geq$$ $$\geq4\sqrt{xy}(x+y+z)+(x-z)(y-z)\geq8xy+(x-z)(y-z).$$ Id est, it's enough to prove that $$\sum_{cyc}\left(\tfrac{4x(x+y+z)+(y-z)^2}{(2x+y+z)^2}+\tfrac{16xy+2(x-z)(y-z)}{(2x+y+z)(2y+x+z)}\right)\geq4$$ or $$\sum_{sym}\left(2x^6+12x^5y-3x^4y^2-13x^3y^3+48x^4yz+218x^3y^2z+\frac{235}{3}x^2y^2z^2\right)\geq0$$ or $$4\sum_{cyc}(x^6-x^4y^2-x^4z^2+x^2y^2z^2)+$$ $$+\sum_{sym}(12x^5y+x^4y^2-13x^3y^3)+\sum_{sym}\left(48x^4yz+218x^3y^2z+\frac{229}{3}x^2y^2z^2\right)\geq0,$$ which is true by Schur and Muirhead.
Let $a=b$ and $c\rightarrow0^+$.
Thus, $\sum\limits_{cyc}\frac{m_a}{b+c}\rightarrow\frac{3}{2},$ which says that your estimation is the best.
Done!