Mathematics Asked by Daniel Kawai on December 22, 2020

**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 (*)?

There is also the following way.

We need to prove that: $$sum_{cyc}frac{sqrt{(2a^2+2b^2-c^2)(2a^2+2c^2-b^2)}}{b^2+c^2}geqfrac{9}{2}.$$ Now, by Holder $$left(sum_{cyc}tfrac{sqrt{(2a^2+2b^2-c^2)(2a^2+2c^2-b^2)}}{b^2+c^2}right)^2sum_{cyc}(2a^2+2b^2-c^2)^2(2a^2+2c^2-b^2)^2(b^2+c^2)^2geq$$ $$geqleft(sum_{cyc}(2a^2+2b^2-c^2)(2a^2+2c^2-b^2)right)^3.$$ Thus, it's enough to prove that: $$4left(sum_{cyc}(2a^2+2b^2-c^2)(2a^2+2c^2-b^2)right)^3geq$$ $$geq81sum_{cyc}(2a^2+2b^2-c^2)^2(2a^2+2c^2-b^2)^2(b^2+c^2)^2$$ or $$36left(sum_{cyc}a^2b^2right)^3geqsum_{cyc}(2a^2+2b^2-c^2)^2(2a^2+2c^2-b^2)^2(b^2+c^2)^2.$$ 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 $$36left(sum_{cyc}(x^2+3xy)right)^3geqsum_{cyc}(x+y+4z)^2(x+z+4y)^2(2x+y+z)^2.$$ 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: $$36(9u^2+3v^2)^3geqsum_{cyc}(3u+3z)^2(3u+3y)^2(3u+x)^2$$ 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 $$(2x+1)(x+5)^2(x-1)^2geq0,$$ which is true because for $y=z=1$ we have $$xy+xz+yz=2x+1>0.$$

Correct answer by Michael Rozenberg on December 22, 2020

Also, we can use SOS here.

Indeed, by your work we need to prove for any triangle that: $$sum_{cyc}frac{ab}{a^2+b^2+4c^2}geqfrac{1}{2}$$ 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 $xgeq ygeq z$.

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

Answered by Michael Rozenberg on December 22, 2020

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