A triangle has sides $a$, $b$, $c$ and medians $m_a$, $m_b$, $m_c$. Show...
$begingroup$
Let $triangle ABC$ have sides $BC=a$, $CA=b$, and $AB=c$. Let $m_a$, $m_b$, $m_c$ be the medians to $BC$, $CA$, and $AB$, respectively. Prove that
$$(ab+bc+ca)left(frac{1}{a}+frac{1}{b}+frac{1}{c}right)geq 2sqrt{3}(m_a+m_b+m_c)$$
My trying: $$Leftrightarrow (ab+bc+ca)left(frac{1}{a}+frac{1}{b}+frac{1}{c}right)geq 3sqrt{3}sqrt{a^2+b^2+c^2}$$
Because: $$m_a+m_b+m_cleq sqrt{3(m_{a}^{2}+m_{b}^{2}+m_{c}^{2})}=frac{3}{2}sqrt{a^2+b^2+c^2}$$
$$Leftrightarrow left(p^2+4Rr+r^2right)left(frac{4R+r}{pr}
right)geq 3sqrt{3}sqrt{2p^2-8Rr-2r^2}$$
I square it, but that doesn't help.
geometry inequality triangle geometric-inequalities sos
edited Dec 30 '18 at 1:59
Michael Rozenberg
105k1892197
asked Dec 16 '18 at 12:14
Winter In My Heart Winter In My Heart
463
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add a comment |
$begingroup$
Let $triangle ABC$ have sides $BC=a$, $CA=b$, and $AB=c$. Let $m_a$, $m_b$, $m_c$ be the medians to $BC$, $CA$, and $AB$, respectively. Prove that
$$(ab+bc+ca)left(frac{1}{a}+frac{1}{b}+frac{1}{c}right)geq 2sqrt{3}(m_a+m_b+m_c)$$
My trying: $$Leftrightarrow (ab+bc+ca)left(frac{1}{a}+frac{1}{b}+frac{1}{c}right)geq 3sqrt{3}sqrt{a^2+b^2+c^2}$$
Because: $$m_a+m_b+m_cleq sqrt{3(m_{a}^{2}+m_{b}^{2}+m_{c}^{2})}=frac{3}{2}sqrt{a^2+b^2+c^2}$$
$$Leftrightarrow left(p^2+4Rr+r^2right)left(frac{4R+r}{pr}
right)geq 3sqrt{3}sqrt{2p^2-8Rr-2r^2}$$
I square it, but that doesn't help.
geometry inequality triangle geometric-inequalities sos
edited Dec 30 '18 at 1:59
Michael Rozenberg
105k1892197
asked Dec 16 '18 at 12:14
Winter In My Heart Winter In My Heart
463
$endgroup$
add a comment |
$begingroup$
Let $triangle ABC$ have sides $BC=a$, $CA=b$, and $AB=c$. Let $m_a$, $m_b$, $m_c$ be the medians to $BC$, $CA$, and $AB$, respectively. Prove that
$$(ab+bc+ca)left(frac{1}{a}+frac{1}{b}+frac{1}{c}right)geq 2sqrt{3}(m_a+m_b+m_c)$$
My trying: $$Leftrightarrow (ab+bc+ca)left(frac{1}{a}+frac{1}{b}+frac{1}{c}right)geq 3sqrt{3}sqrt{a^2+b^2+c^2}$$
Because: $$m_a+m_b+m_cleq sqrt{3(m_{a}^{2}+m_{b}^{2}+m_{c}^{2})}=frac{3}{2}sqrt{a^2+b^2+c^2}$$
$$Leftrightarrow left(p^2+4Rr+r^2right)left(frac{4R+r}{pr}
right)geq 3sqrt{3}sqrt{2p^2-8Rr-2r^2}$$
I square it, but that doesn't help.
geometry inequality triangle geometric-inequalities sos
edited Dec 30 '18 at 1:59
Michael Rozenberg
105k1892197
asked Dec 16 '18 at 12:14
Winter In My Heart Winter In My Heart
463
$endgroup$
Let $triangle ABC$ have sides $BC=a$, $CA=b$, and $AB=c$. Let $m_a$, $m_b$, $m_c$ be the medians to $BC$, $CA$, and $AB$, respectively. Prove that
$$(ab+bc+ca)left(frac{1}{a}+frac{1}{b}+frac{1}{c}right)geq 2sqrt{3}(m_a+m_b+m_c)$$
My trying: $$Leftrightarrow (ab+bc+ca)left(frac{1}{a}+frac{1}{b}+frac{1}{c}right)geq 3sqrt{3}sqrt{a^2+b^2+c^2}$$
Because: $$m_a+m_b+m_cleq sqrt{3(m_{a}^{2}+m_{b}^{2}+m_{c}^{2})}=frac{3}{2}sqrt{a^2+b^2+c^2}$$
$$Leftrightarrow left(p^2+4Rr+r^2right)left(frac{4R+r}{pr}
right)geq 3sqrt{3}sqrt{2p^2-8Rr-2r^2}$$
I square it, but that doesn't help.
geometry inequality triangle geometric-inequalities sos
geometry inequality triangle geometric-inequalities sos
edited Dec 30 '18 at 1:59
Michael Rozenberg
105k1892197
asked Dec 16 '18 at 12:14
Winter In My Heart Winter In My Heart
463
edited Dec 30 '18 at 1:59
Michael Rozenberg
105k1892197
asked Dec 16 '18 at 12:14
Winter In My Heart Winter In My Heart
463
edited Dec 30 '18 at 1:59
Michael Rozenberg
105k1892197
edited Dec 30 '18 at 1:59
Michael Rozenberg
105k1892197
edited Dec 30 '18 at 1:59
Michael Rozenberg
105k1892197
105k1892197
asked Dec 16 '18 at 12:14
Winter In My Heart Winter In My Heart
463
asked Dec 16 '18 at 12:14
Winter In My Heart Winter In My Heart
463
asked Dec 16 '18 at 12:14
Winter In My Heart Winter In My Heart
463
463
add a comment |
add a comment |
1 Answer
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oldest
votes
$begingroup$
The hint:
Draw the $Delta ABC$ with medians $m_b=BD$ and $CE=m_a$.
Thus, by the Ptolemy's theorem for the equilateral $BCDE$ we obtain:
$$BEcdot DC+BCcdot EDgeq BDcdot CE$$ or
$$frac{bc}{4}+frac{a^2}{2}geq m_am_b.$$
Hence, $$sum_{cyc}m_a=sqrt{sum_{cyc}left(m_a^2+2m_bm_cright)}leqsqrt{sum_{cyc}left(m_a^2+2left(frac{bc}{4}+frac{a^2}{2}right)right)}=$$
$$=sqrt{sum_{cyc}left(frac{1}{4}(2b^2+2c^2-a^2)+2left(frac{bc}{4}+frac{a^2}{2}right)right)}=frac{1}{2}sqrt{sum_{cyc}(7a^2+2ab)}.$$
It est, it's enough to prove that
$$frac{(ab+ac+bc)^4}{3a^2b^2c^2}geqsum_{cyc}(7a^2+2ab).$$
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
$$left(sum_{cyc}(x^2+3xy)right)^4geq3prod_{cyc}(x+y)^2sum_{cyc}(14x^2+14y^2+2x^2+6xy)$$ or
$$left(sum_{cyc}(x^2+3xy)right)^4geq12prod_{cyc}(x+y)^2sum_{cyc}(4x^2+5xy)$$ or
$$sum_{cyc}(x^8+12x^7y+12x^7z+10x^6y^2+10x^6z^2-12x^5y^3-12x^5z^3-21x^4y^4)+$$
$$+24xyzsum_{cyc}(x^5-x^3y^2-x^3z^2+x^2y^2z)geq0.$$
Now, $$sum_{cyc}(x^8+12x^7y+12x^7z+10x^6y^2+10x^6z^2-12x^5y^3-12x^5z^3-21x^4y^4)=$$
$$=frac{1}{2}sum_{cyc}(x^8+y^8+24x^7y+24xy^7+20x^6y^2+20x^2y^6-24x^5y^3-24x^3y^5-42x^4y^4)=$$
$$=frac{1}{2}sum_{cyc}(x^2-y^2)^2(x^4+24x^3y+22x^2y^2+24xy^3+y^4)geq0$$ and let $xgeq ygeq z$.
Thus,
$$sum_{cyc}(x^5-x^3y^2-x^3z^2+x^2y^2z)=$$
$$=frac{1}{2}sum_{cyc}(x^5-x^3y^2-x^2y^3+y^5-(x^3z^2-x^2yz^2-y^2xz^2+y^3z^2))=$$
$$=frac{1}{2}sum_{cyc}(x-y)^2((x+y)(x^2+xy+y^2)-(x+y)z^2)=$$
$$=frac{1}{2}sum_{cyc}(x-y)^2(x+y)(x^2+xy+y^2-z^2)geq$$
$$geqfrac{1}{2}left((x-z)^2(x+z)(x^2-y^2)+(y-z)^2(y+z)(y^2-x^2)right)geq$$
$$geqfrac{1}{2}left((y-z)^2(x+z)(x^2-y^2)+(y-z)^2(y+z)(y^2-x^2)right)=frac{1}{2}(y-z)^2(x-y)^2(x+y)geq0.$$
Done!
answered Dec 16 '18 at 12:26
Michael RozenbergMichael Rozenberg
105k1892197
$endgroup$
$begingroup$
Please help me prove the last. Do you have another solution?
$endgroup$
– Winter In My Heart
Dec 16 '18 at 12:29
$begingroup$
@Winter In My Heart Also, it's true after the Ravi's substitution and full expanding. It's very ugly, but it's proof.
$endgroup$
– Michael Rozenberg
Dec 16 '18 at 12:47
$begingroup$
No problem, show it please
$endgroup$
– Winter In My Heart
Dec 16 '18 at 12:51
$begingroup$
@Winter In My Heart I fixed a typo in my proof. See now.
$endgroup$
– Michael Rozenberg
Dec 17 '18 at 6:45
add a comment |
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$begingroup$
The hint:
Draw the $Delta ABC$ with medians $m_b=BD$ and $CE=m_a$.
Thus, by the Ptolemy's theorem for the equilateral $BCDE$ we obtain:
$$BEcdot DC+BCcdot EDgeq BDcdot CE$$ or
$$frac{bc}{4}+frac{a^2}{2}geq m_am_b.$$
Hence, $$sum_{cyc}m_a=sqrt{sum_{cyc}left(m_a^2+2m_bm_cright)}leqsqrt{sum_{cyc}left(m_a^2+2left(frac{bc}{4}+frac{a^2}{2}right)right)}=$$
$$=sqrt{sum_{cyc}left(frac{1}{4}(2b^2+2c^2-a^2)+2left(frac{bc}{4}+frac{a^2}{2}right)right)}=frac{1}{2}sqrt{sum_{cyc}(7a^2+2ab)}.$$
It est, it's enough to prove that
$$frac{(ab+ac+bc)^4}{3a^2b^2c^2}geqsum_{cyc}(7a^2+2ab).$$
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
$$left(sum_{cyc}(x^2+3xy)right)^4geq3prod_{cyc}(x+y)^2sum_{cyc}(14x^2+14y^2+2x^2+6xy)$$ or
$$left(sum_{cyc}(x^2+3xy)right)^4geq12prod_{cyc}(x+y)^2sum_{cyc}(4x^2+5xy)$$ or
$$sum_{cyc}(x^8+12x^7y+12x^7z+10x^6y^2+10x^6z^2-12x^5y^3-12x^5z^3-21x^4y^4)+$$
$$+24xyzsum_{cyc}(x^5-x^3y^2-x^3z^2+x^2y^2z)geq0.$$
Now, $$sum_{cyc}(x^8+12x^7y+12x^7z+10x^6y^2+10x^6z^2-12x^5y^3-12x^5z^3-21x^4y^4)=$$
$$=frac{1}{2}sum_{cyc}(x^8+y^8+24x^7y+24xy^7+20x^6y^2+20x^2y^6-24x^5y^3-24x^3y^5-42x^4y^4)=$$
$$=frac{1}{2}sum_{cyc}(x^2-y^2)^2(x^4+24x^3y+22x^2y^2+24xy^3+y^4)geq0$$ and let $xgeq ygeq z$.
Thus,
$$sum_{cyc}(x^5-x^3y^2-x^3z^2+x^2y^2z)=$$
$$=frac{1}{2}sum_{cyc}(x^5-x^3y^2-x^2y^3+y^5-(x^3z^2-x^2yz^2-y^2xz^2+y^3z^2))=$$
$$=frac{1}{2}sum_{cyc}(x-y)^2((x+y)(x^2+xy+y^2)-(x+y)z^2)=$$
$$=frac{1}{2}sum_{cyc}(x-y)^2(x+y)(x^2+xy+y^2-z^2)geq$$
$$geqfrac{1}{2}left((x-z)^2(x+z)(x^2-y^2)+(y-z)^2(y+z)(y^2-x^2)right)geq$$
$$geqfrac{1}{2}left((y-z)^2(x+z)(x^2-y^2)+(y-z)^2(y+z)(y^2-x^2)right)=frac{1}{2}(y-z)^2(x-y)^2(x+y)geq0.$$
Done!
answered Dec 16 '18 at 12:26
Michael RozenbergMichael Rozenberg
105k1892197
$endgroup$
$begingroup$
Please help me prove the last. Do you have another solution?
$endgroup$
– Winter In My Heart
Dec 16 '18 at 12:29
$begingroup$
@Winter In My Heart Also, it's true after the Ravi's substitution and full expanding. It's very ugly, but it's proof.
$endgroup$
– Michael Rozenberg
Dec 16 '18 at 12:47
$begingroup$
No problem, show it please
$endgroup$
– Winter In My Heart
Dec 16 '18 at 12:51
$begingroup$
@Winter In My Heart I fixed a typo in my proof. See now.
$endgroup$
– Michael Rozenberg
Dec 17 '18 at 6:45
add a comment |
$begingroup$
The hint:
Draw the $Delta ABC$ with medians $m_b=BD$ and $CE=m_a$.
Thus, by the Ptolemy's theorem for the equilateral $BCDE$ we obtain:
$$BEcdot DC+BCcdot EDgeq BDcdot CE$$ or
$$frac{bc}{4}+frac{a^2}{2}geq m_am_b.$$
Hence, $$sum_{cyc}m_a=sqrt{sum_{cyc}left(m_a^2+2m_bm_cright)}leqsqrt{sum_{cyc}left(m_a^2+2left(frac{bc}{4}+frac{a^2}{2}right)right)}=$$
$$=sqrt{sum_{cyc}left(frac{1}{4}(2b^2+2c^2-a^2)+2left(frac{bc}{4}+frac{a^2}{2}right)right)}=frac{1}{2}sqrt{sum_{cyc}(7a^2+2ab)}.$$
It est, it's enough to prove that
$$frac{(ab+ac+bc)^4}{3a^2b^2c^2}geqsum_{cyc}(7a^2+2ab).$$
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
$$left(sum_{cyc}(x^2+3xy)right)^4geq3prod_{cyc}(x+y)^2sum_{cyc}(14x^2+14y^2+2x^2+6xy)$$ or
$$left(sum_{cyc}(x^2+3xy)right)^4geq12prod_{cyc}(x+y)^2sum_{cyc}(4x^2+5xy)$$ or
$$sum_{cyc}(x^8+12x^7y+12x^7z+10x^6y^2+10x^6z^2-12x^5y^3-12x^5z^3-21x^4y^4)+$$
$$+24xyzsum_{cyc}(x^5-x^3y^2-x^3z^2+x^2y^2z)geq0.$$
Now, $$sum_{cyc}(x^8+12x^7y+12x^7z+10x^6y^2+10x^6z^2-12x^5y^3-12x^5z^3-21x^4y^4)=$$
$$=frac{1}{2}sum_{cyc}(x^8+y^8+24x^7y+24xy^7+20x^6y^2+20x^2y^6-24x^5y^3-24x^3y^5-42x^4y^4)=$$
$$=frac{1}{2}sum_{cyc}(x^2-y^2)^2(x^4+24x^3y+22x^2y^2+24xy^3+y^4)geq0$$ and let $xgeq ygeq z$.
Thus,
$$sum_{cyc}(x^5-x^3y^2-x^3z^2+x^2y^2z)=$$
$$=frac{1}{2}sum_{cyc}(x^5-x^3y^2-x^2y^3+y^5-(x^3z^2-x^2yz^2-y^2xz^2+y^3z^2))=$$
$$=frac{1}{2}sum_{cyc}(x-y)^2((x+y)(x^2+xy+y^2)-(x+y)z^2)=$$
$$=frac{1}{2}sum_{cyc}(x-y)^2(x+y)(x^2+xy+y^2-z^2)geq$$
$$geqfrac{1}{2}left((x-z)^2(x+z)(x^2-y^2)+(y-z)^2(y+z)(y^2-x^2)right)geq$$
$$geqfrac{1}{2}left((y-z)^2(x+z)(x^2-y^2)+(y-z)^2(y+z)(y^2-x^2)right)=frac{1}{2}(y-z)^2(x-y)^2(x+y)geq0.$$
Done!
answered Dec 16 '18 at 12:26
Michael RozenbergMichael Rozenberg
105k1892197
$endgroup$
$begingroup$
Please help me prove the last. Do you have another solution?
$endgroup$
– Winter In My Heart
Dec 16 '18 at 12:29
$begingroup$
@Winter In My Heart Also, it's true after the Ravi's substitution and full expanding. It's very ugly, but it's proof.
$endgroup$
– Michael Rozenberg
Dec 16 '18 at 12:47
$begingroup$
No problem, show it please
$endgroup$
– Winter In My Heart
Dec 16 '18 at 12:51
$begingroup$
@Winter In My Heart I fixed a typo in my proof. See now.
$endgroup$
– Michael Rozenberg
Dec 17 '18 at 6:45
add a comment |
$begingroup$
The hint:
Draw the $Delta ABC$ with medians $m_b=BD$ and $CE=m_a$.
Thus, by the Ptolemy's theorem for the equilateral $BCDE$ we obtain:
$$BEcdot DC+BCcdot EDgeq BDcdot CE$$ or
$$frac{bc}{4}+frac{a^2}{2}geq m_am_b.$$
Hence, $$sum_{cyc}m_a=sqrt{sum_{cyc}left(m_a^2+2m_bm_cright)}leqsqrt{sum_{cyc}left(m_a^2+2left(frac{bc}{4}+frac{a^2}{2}right)right)}=$$
$$=sqrt{sum_{cyc}left(frac{1}{4}(2b^2+2c^2-a^2)+2left(frac{bc}{4}+frac{a^2}{2}right)right)}=frac{1}{2}sqrt{sum_{cyc}(7a^2+2ab)}.$$
It est, it's enough to prove that
$$frac{(ab+ac+bc)^4}{3a^2b^2c^2}geqsum_{cyc}(7a^2+2ab).$$
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
$$left(sum_{cyc}(x^2+3xy)right)^4geq3prod_{cyc}(x+y)^2sum_{cyc}(14x^2+14y^2+2x^2+6xy)$$ or
$$left(sum_{cyc}(x^2+3xy)right)^4geq12prod_{cyc}(x+y)^2sum_{cyc}(4x^2+5xy)$$ or
$$sum_{cyc}(x^8+12x^7y+12x^7z+10x^6y^2+10x^6z^2-12x^5y^3-12x^5z^3-21x^4y^4)+$$
$$+24xyzsum_{cyc}(x^5-x^3y^2-x^3z^2+x^2y^2z)geq0.$$
Now, $$sum_{cyc}(x^8+12x^7y+12x^7z+10x^6y^2+10x^6z^2-12x^5y^3-12x^5z^3-21x^4y^4)=$$
$$=frac{1}{2}sum_{cyc}(x^8+y^8+24x^7y+24xy^7+20x^6y^2+20x^2y^6-24x^5y^3-24x^3y^5-42x^4y^4)=$$
$$=frac{1}{2}sum_{cyc}(x^2-y^2)^2(x^4+24x^3y+22x^2y^2+24xy^3+y^4)geq0$$ and let $xgeq ygeq z$.
Thus,
$$sum_{cyc}(x^5-x^3y^2-x^3z^2+x^2y^2z)=$$
$$=frac{1}{2}sum_{cyc}(x^5-x^3y^2-x^2y^3+y^5-(x^3z^2-x^2yz^2-y^2xz^2+y^3z^2))=$$
$$=frac{1}{2}sum_{cyc}(x-y)^2((x+y)(x^2+xy+y^2)-(x+y)z^2)=$$
$$=frac{1}{2}sum_{cyc}(x-y)^2(x+y)(x^2+xy+y^2-z^2)geq$$
$$geqfrac{1}{2}left((x-z)^2(x+z)(x^2-y^2)+(y-z)^2(y+z)(y^2-x^2)right)geq$$
$$geqfrac{1}{2}left((y-z)^2(x+z)(x^2-y^2)+(y-z)^2(y+z)(y^2-x^2)right)=frac{1}{2}(y-z)^2(x-y)^2(x+y)geq0.$$
Done!
answered Dec 16 '18 at 12:26
Michael RozenbergMichael Rozenberg
105k1892197
$endgroup$
The hint:
Draw the $Delta ABC$ with medians $m_b=BD$ and $CE=m_a$.
Thus, by the Ptolemy's theorem for the equilateral $BCDE$ we obtain:
$$BEcdot DC+BCcdot EDgeq BDcdot CE$$ or
$$frac{bc}{4}+frac{a^2}{2}geq m_am_b.$$
Hence, $$sum_{cyc}m_a=sqrt{sum_{cyc}left(m_a^2+2m_bm_cright)}leqsqrt{sum_{cyc}left(m_a^2+2left(frac{bc}{4}+frac{a^2}{2}right)right)}=$$
$$=sqrt{sum_{cyc}left(frac{1}{4}(2b^2+2c^2-a^2)+2left(frac{bc}{4}+frac{a^2}{2}right)right)}=frac{1}{2}sqrt{sum_{cyc}(7a^2+2ab)}.$$
It est, it's enough to prove that
$$frac{(ab+ac+bc)^4}{3a^2b^2c^2}geqsum_{cyc}(7a^2+2ab).$$
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
$$left(sum_{cyc}(x^2+3xy)right)^4geq3prod_{cyc}(x+y)^2sum_{cyc}(14x^2+14y^2+2x^2+6xy)$$ or
$$left(sum_{cyc}(x^2+3xy)right)^4geq12prod_{cyc}(x+y)^2sum_{cyc}(4x^2+5xy)$$ or
$$sum_{cyc}(x^8+12x^7y+12x^7z+10x^6y^2+10x^6z^2-12x^5y^3-12x^5z^3-21x^4y^4)+$$
$$+24xyzsum_{cyc}(x^5-x^3y^2-x^3z^2+x^2y^2z)geq0.$$
Now, $$sum_{cyc}(x^8+12x^7y+12x^7z+10x^6y^2+10x^6z^2-12x^5y^3-12x^5z^3-21x^4y^4)=$$
$$=frac{1}{2}sum_{cyc}(x^8+y^8+24x^7y+24xy^7+20x^6y^2+20x^2y^6-24x^5y^3-24x^3y^5-42x^4y^4)=$$
$$=frac{1}{2}sum_{cyc}(x^2-y^2)^2(x^4+24x^3y+22x^2y^2+24xy^3+y^4)geq0$$ and let $xgeq ygeq z$.
Thus,
$$sum_{cyc}(x^5-x^3y^2-x^3z^2+x^2y^2z)=$$
$$=frac{1}{2}sum_{cyc}(x^5-x^3y^2-x^2y^3+y^5-(x^3z^2-x^2yz^2-y^2xz^2+y^3z^2))=$$
$$=frac{1}{2}sum_{cyc}(x-y)^2((x+y)(x^2+xy+y^2)-(x+y)z^2)=$$
$$=frac{1}{2}sum_{cyc}(x-y)^2(x+y)(x^2+xy+y^2-z^2)geq$$
$$geqfrac{1}{2}left((x-z)^2(x+z)(x^2-y^2)+(y-z)^2(y+z)(y^2-x^2)right)geq$$
$$geqfrac{1}{2}left((y-z)^2(x+z)(x^2-y^2)+(y-z)^2(y+z)(y^2-x^2)right)=frac{1}{2}(y-z)^2(x-y)^2(x+y)geq0.$$
Done!
answered Dec 16 '18 at 12:26
Michael RozenbergMichael Rozenberg
105k1892197
edited Dec 17 '18 at 6:44
answered Dec 16 '18 at 12:26
Michael RozenbergMichael Rozenberg
105k1892197
answered Dec 16 '18 at 12:26
Michael RozenbergMichael Rozenberg
105k1892197
answered Dec 16 '18 at 12:26
Michael RozenbergMichael Rozenberg
105k1892197
105k1892197
$begingroup$
Please help me prove the last. Do you have another solution?
$endgroup$
– Winter In My Heart
Dec 16 '18 at 12:29
$begingroup$
@Winter In My Heart Also, it's true after the Ravi's substitution and full expanding. It's very ugly, but it's proof.
$endgroup$
– Michael Rozenberg
Dec 16 '18 at 12:47
$begingroup$
No problem, show it please
$endgroup$
– Winter In My Heart
Dec 16 '18 at 12:51
$begingroup$
@Winter In My Heart I fixed a typo in my proof. See now.
$endgroup$
– Michael Rozenberg
Dec 17 '18 at 6:45
add a comment |
$begingroup$
Please help me prove the last. Do you have another solution?
$endgroup$
– Winter In My Heart
Dec 16 '18 at 12:29
$begingroup$
@Winter In My Heart Also, it's true after the Ravi's substitution and full expanding. It's very ugly, but it's proof.
$endgroup$
– Michael Rozenberg
Dec 16 '18 at 12:47
$begingroup$
No problem, show it please
$endgroup$
– Winter In My Heart
Dec 16 '18 at 12:51
$begingroup$
@Winter In My Heart I fixed a typo in my proof. See now.
$endgroup$
– Michael Rozenberg
Dec 17 '18 at 6:45
$begingroup$
Please help me prove the last. Do you have another solution?
$endgroup$
– Winter In My Heart
Dec 16 '18 at 12:29
$begingroup$
Please help me prove the last. Do you have another solution?
$endgroup$
– Winter In My Heart
Dec 16 '18 at 12:29
$begingroup$
@Winter In My Heart Also, it's true after the Ravi's substitution and full expanding. It's very ugly, but it's proof.
$endgroup$
– Michael Rozenberg
Dec 16 '18 at 12:47
$begingroup$
@Winter In My Heart Also, it's true after the Ravi's substitution and full expanding. It's very ugly, but it's proof.
$endgroup$
– Michael Rozenberg
Dec 16 '18 at 12:47
$begingroup$
No problem, show it please
$endgroup$
– Winter In My Heart
Dec 16 '18 at 12:51
$begingroup$
No problem, show it please
$endgroup$
– Winter In My Heart
Dec 16 '18 at 12:51
$begingroup$
@Winter In My Heart I fixed a typo in my proof. See now.
$endgroup$
– Michael Rozenberg
Dec 17 '18 at 6:45
$begingroup$
@Winter In My Heart I fixed a typo in my proof. See now.
$endgroup$
– Michael Rozenberg
Dec 17 '18 at 6:45
add a comment |
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