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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:

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

My solution

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

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

And also:

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

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:

$$m_a=y+z,\quad\quad m_b=z+x,\quad\quad m_c=x+y$$

So, because of:

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

we want to prove that:

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

or equivalently:

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

If we clear the denominators and develop everything, then:

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

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

and:

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

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

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:

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

My question

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

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2 Answers 2

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There is also the following way.

We need to prove that: $$\sum_{cyc}\frac{\sqrt{(2a^2+2b^2-c^2)(2a^2+2c^2-b^2)}}{b^2+c^2}\geq\frac{9}{2}.$$ Now, by Holder $$\left(\sum_{cyc}\tfrac{\sqrt{(2a^2+2b^2-c^2)(2a^2+2c^2-b^2)}}{b^2+c^2}\right)^2\sum_{cyc}(2a^2+2b^2-c^2)^2(2a^2+2c^2-b^2)^2(b^2+c^2)^2\geq$$ $$\geq\left(\sum_{cyc}(2a^2+2b^2-c^2)(2a^2+2c^2-b^2)\right)^3.$$ Thus, it's enough to prove that: $$4\left(\sum_{cyc}(2a^2+2b^2-c^2)(2a^2+2c^2-b^2)\right)^3\geq$$ $$\geq81\sum_{cyc}(2a^2+2b^2-c^2)^2(2a^2+2c^2-b^2)^2(b^2+c^2)^2$$ or $$36\left(\sum_{cyc}a^2b^2\right)^3\geq\sum_{cyc}(2a^2+2b^2-c^2)^2(2a^2+2c^2-b^2)^2(b^2+c^2)^2.$$ Now, let $b^2+c^2-a^2=x$, $a^2+c^2-b^2=y$ and $a^2+b^2-c^2=z$.

Thus, we need to prove that $$36\left(\sum_{cyc}(x^2+3xy)\right)^3\geq\sum_{cyc}(x+y+4z)^2(x+z+4y)^2(2x+y+z)^2.$$ Now, let $x+y+z=3u$, $xy+xz+yz=3v^2$ and $xyz=w^3$.

We see that $$\sum_{cyc}xy=\sum_{cyc}(b^2+c^2-a^2)(a^2+c^2-b^2)=\sum_{cyc}(2a^2b^2-a^4)=16S^2>0$$ and we need to prove that: $$36(9u^2+3v^2)^3\geq\sum_{cyc}(3u+3z)^2(3u+3y)^2(3u+x)^2$$ or $f(w^3)\geq0$, where $f$ is a concave function because the coefficient before $w^6$ is negative.

But the concave function gets a minimal value for an extreme value of $w^3$,

which happens for equality case of two variables.

Since our inequality is homogeneous and symmetric, it's enough to assume $y=z=1$

(the case $y=z=0$ is impossible), which gives $$(2x+1)(x+5)^2(x-1)^2\geq0,$$ which is true because for $y=z=1$ we have $$xy+xz+yz=2x+1>0.$$

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Also, we can use SOS here.

Indeed, by your work we need to prove for any triangle that: $$\sum_{cyc}\frac{ab}{a^2+b^2+4c^2}\geq\frac{1}{2}$$ or $$\sum_{cyc}\left(\frac{ab}{a^2+b^2+4c^2}-\frac{1}{6}\right)\geq0$$ or $$\sum_{cyc}\frac{6ab-a^2-b^2-4c^2}{a^2+b^2+4c^2}\geq0$$ or $$\sum_{cyc}\frac{(b-c)(3a-b+2c)-(c-a)(3b-a+2c)}{a^2+b^2+4c^2}\geq0$$ or $$\sum_{cyc}(a-b)\left(\frac{3c-a+2b}{a^2+c^2+4b^2}-\frac{3c-b+2a}{b^2+c^2+4a^2}\right)\geq0$$ or $$\sum_{cyc}(a-b)^2(-2a^2-2b^2-c^2+ab+3ac+3bc)(a^2+b^2+4c^2)\geq0.$$ Now, let $a=y+z,$ $b=x+z$ and $c=x+y.$

Thus, $x$, $y$ and $z$ are positives and we need to prove that $$\sum_{cyc}(x-y)^2(5xy+3xz+3yz-3z^2)(a^2+b^2+4c^2)\geq0,$$ for which it's enough to prove that: $$\sum_{cyc}(x-y)^2z(x+y-z)(a^2+b^2+4c^2)\geq0.$$ Now, let $x\geq y\geq z$.

Thus, $$y\sum_{cyc}(x-y)^2z(x+y-z)(a^2+b^2+4c^2)\geq$$ $$\geq y^2(x-z)^2(x+z-y)(a^2+c^2+4b^2)+y(y-z)^2x(y+z-x)(b^2+c^2+4a^2)\geq$$ $$\geq x^2(y-z)^2(x-y)(a^2+c^2+4b^2)+y(y-z)^2x(y-x)(b^2+c^2+4a^2)=$$ $$=x(x-y)(y-z)^2(x(a^2+c^2+4b^2)-y(b^2+c^2+4a^2))=$$ $$=\frac{1}{2}x(x-y)(y-z)^2((b+c-a)(a^2+c^2+4b^2)-(a+c-b)(b^2+c^2+4a^2))=$$ $$=\frac{1}{2}x(x-y)(y-z)^2(b-a)(5a^2+5b^2+2c^2+3ac+3bc)=$$ $$=\frac{1}{2}x(x-y)^2(y-z)^2(5a^2+5b^2+2c^2+3ac+3bc)\geq0$$ and we are done!

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