Inequality involving medians

Problem

The sides of a triangle are $a$, $b$ and $c$ and the lengths of the corresponding medians are $m_a$, $m_b$ and $m_c$. I want to prove that:

$$frac{m_am_b}{a^2+b^2}+frac{m_bm_c}{b^2+c^2}+frac{m_cm_a}{c^2+a^2}geqfrac{9}{8}.$$

My solution

We can calculate the medians in terms of the sides of the triangle:

$$m_a^2=frac{1}{4}(-a^2+2b^2+2c^2),quadquad m_b^2=frac{1}{4}(2a^2-b^2+2c^2),quadquad m_c^2=frac{1}{4}(2a^2+2b^2-c^2)$$

And also:

$$a^2=frac{4}{9}(-m_a^2+2m_b^2+2m_c^2),quadquad b^2=frac{4}{9}(2m_a^2-m_b^2+2m_c^2),quadquad c^2=frac{4}{9}(2m_a^2+2m_b^2-m_c^2)$$

Moreover, it is possible to prove that $m_a$, $m_b$ and $m_c$ are sides of another triangle.

Indeed, let $ABC$ be a triangle such that $BC=a$, $CA=b$ and $AB=c$. Let $D$, $E$ and $F$ be the midpoints of $BC$, $CA$ and $AB$. Let the line $EF$ and the line $l$ parallel to $AB$ passing through $C$ meet at $X$. Then $CDEX$ and $AFCX$ are parallelograms, and thus $AD=m_a$, $DX=BE=m_b$ and $XA=CF=m_c$ are sides of a triangle.

Also, if the numbers $m_a$, $m_b$ and $m_c$ are sides of a triangle, then the numbers $a$, $b$ and $c$ so defined are sides of a triangle.

Therefore, the numbers $a$, $b$ and $c$ are sides of a triangle if and only if the numbers $m_a$, $m_b$ and $m_c$ are sides of a triangle. And it is equivalent to the existence of positive real numbers $x$, $y$ and $z$ such that:

$$m_a=y+z,quadquad m_b=z+x,quadquad m_c=x+y$$

So, because of:

$$a^2+b^2=frac{4}{9}(m_a^2+m_b^2+4m_c^2),quadquad b^2+c^2=frac{4}{9}(4m_a^2+m_b^2+m_c^2),quadquad c^2+a^2=frac{4}{9}(m_a^2+4m_b^2+m_c^2)$$

we want to prove that:

$$frac{m_am_b}{m_a^2+m_b^2+4m_c^2}+frac{m_bm_c}{4m_a^2+m_b^2+m_c^2}+frac{m_cm_a}{m_a^2+4m_b^2+m_c^2}geqfrac{1}{2},$$

or equivalently:

$$tag{*}frac{(x+y)(x+z)}{(x+y)^2+(x+z)^2+4(y+z)^2}+frac{(x+y)(y+z)}{(x+y)^2+4(x+z)^2+(y+z)^2}+frac{(x+z)(y+z)}{4(x+y)^2+(x+z)^2+(y+z)^2}geqfrac{1}{2}.$$

If we clear the denominators and develop everything, then:

$$2sum_{cyc}(x+y)(x+z)left(4(x+y)^2+(x+z)^2+(y+z)^2right)left((x+y)^2+4(x+z)^2+(y+z)^2right)=$$

$$25S_{6,0,0}+190S_{5,1,0}+302S_{4,2,0}+313S_{4,1,1}+187S_{3,3,0}+1038S_{3,2,1}+249S_{2,2,2},$$

and:

$$left(4(x+y)^2+(x+z)^2+(y+z)^2right)left((x+y)^2+4(x+z)^2+(y+z)^2right)left((x+y)^2+(x+z)^2+4(y+z)^2right)=$$

$$25S_{6,0,0}+150S_{5,1,0}+327S_{4,2,0}+288S_{4,1,1}+202S_{3,3,0}+1056S_{3,2,1}+256S_{2,2,2},$$

where:

$$sum_{cyc}f(x,y,z)=f(x,y,z)+f(y,z,x)+f(z,x,y),$$

and:

$$S_{a,b,c}=sum_{sym}x^ay^bz^c=x^ay^bz^c+x^ay^cz^b+x^by^az^c+x^by^cz^a+x^cy^az^b+x^cy^bz^a.$$

Then the inequality is equivalent to:

$$40S_{5,1,0}+25S_{4,1,1}geq25S_{4,2,0}+15S_{3,3,0}+18S_{3,2,1}+7S_{2,2,2},$$

which can be solved easily by Muirhead:

$$25S_{5,1,0}geq25S_{4,2,0},quadquad 15S_{5,1,0}geq15S_{3,3,0},quadquad 18S_{4,1,1}geq18S_{3,2,1},quadquad 7S_{4,1,1}geq7S_{2,2,2}.$$

My question

Is there a shorter and less painful solution without having to clear up denominators and develop everything from (*)?