How to show $7^{th}$ degree polynomial is non-positive in $[0,1]$
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Let $0le xle 1$, show that inequality
$$99x^7-381x^6+225x^5-415x^4+157x^3-3x^2-x-1le 0$$
This problem comes from the fact that I solved a different inequality.I tried to solve it by factorizing it to see if I could get symbols.but I failed.
this inequality is hold by wolfapha Test it.
inequality polynomials a.m.-g.m.-inequality
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add a comment |
$begingroup$
Let $0le xle 1$, show that inequality
$$99x^7-381x^6+225x^5-415x^4+157x^3-3x^2-x-1le 0$$
This problem comes from the fact that I solved a different inequality.I tried to solve it by factorizing it to see if I could get symbols.but I failed.
this inequality is hold by wolfapha Test it.
inequality polynomials a.m.-g.m.-inequality
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Not sure this helps, but here's what the part closest to the $x$-axis looks like for $x < 1$: wolframalpha.com/input/…
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– David K
Dec 19 '18 at 12:44
1
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How about using Sturm's Theorem en.wikipedia.org/wiki/Sturm%27s_theorem and show that the polynomial does not have real roots in the interval (0,1]?
$endgroup$
– Johannes Huisman
Dec 19 '18 at 12:59
add a comment |
$begingroup$
Let $0le xle 1$, show that inequality
$$99x^7-381x^6+225x^5-415x^4+157x^3-3x^2-x-1le 0$$
This problem comes from the fact that I solved a different inequality.I tried to solve it by factorizing it to see if I could get symbols.but I failed.
this inequality is hold by wolfapha Test it.
inequality polynomials a.m.-g.m.-inequality
$endgroup$
Let $0le xle 1$, show that inequality
$$99x^7-381x^6+225x^5-415x^4+157x^3-3x^2-x-1le 0$$
This problem comes from the fact that I solved a different inequality.I tried to solve it by factorizing it to see if I could get symbols.but I failed.
this inequality is hold by wolfapha Test it.
inequality polynomials a.m.-g.m.-inequality
inequality polynomials a.m.-g.m.-inequality
edited Dec 19 '18 at 13:12
Michael Rozenberg
109k1896201
109k1896201
asked Dec 19 '18 at 12:12
inequalityinequality
777522
777522
$begingroup$
Not sure this helps, but here's what the part closest to the $x$-axis looks like for $x < 1$: wolframalpha.com/input/…
$endgroup$
– David K
Dec 19 '18 at 12:44
1
$begingroup$
How about using Sturm's Theorem en.wikipedia.org/wiki/Sturm%27s_theorem and show that the polynomial does not have real roots in the interval (0,1]?
$endgroup$
– Johannes Huisman
Dec 19 '18 at 12:59
add a comment |
$begingroup$
Not sure this helps, but here's what the part closest to the $x$-axis looks like for $x < 1$: wolframalpha.com/input/…
$endgroup$
– David K
Dec 19 '18 at 12:44
1
$begingroup$
How about using Sturm's Theorem en.wikipedia.org/wiki/Sturm%27s_theorem and show that the polynomial does not have real roots in the interval (0,1]?
$endgroup$
– Johannes Huisman
Dec 19 '18 at 12:59
$begingroup$
Not sure this helps, but here's what the part closest to the $x$-axis looks like for $x < 1$: wolframalpha.com/input/…
$endgroup$
– David K
Dec 19 '18 at 12:44
$begingroup$
Not sure this helps, but here's what the part closest to the $x$-axis looks like for $x < 1$: wolframalpha.com/input/…
$endgroup$
– David K
Dec 19 '18 at 12:44
1
1
$begingroup$
How about using Sturm's Theorem en.wikipedia.org/wiki/Sturm%27s_theorem and show that the polynomial does not have real roots in the interval (0,1]?
$endgroup$
– Johannes Huisman
Dec 19 '18 at 12:59
$begingroup$
How about using Sturm's Theorem en.wikipedia.org/wiki/Sturm%27s_theorem and show that the polynomial does not have real roots in the interval (0,1]?
$endgroup$
– Johannes Huisman
Dec 19 '18 at 12:59
add a comment |
1 Answer
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$begingroup$
Because by AM-GM $$1+x+3x^2-157x^3+415x^4-225x^5+381x^6-99x^7=$$
$$=1+x+3x^2-157x^3+370x^4+x^4(45-225x+282x^2)+99x^6(1-x)geq$$
$$geq370x^4+3x^2+x+1-157x^3=6cdotfrac{185}{3}x^4+3x^2+x+1-157x^3ge$$
$$geq9sqrt[9]{left(frac{185}{3}x^4right)^6cdot3x^2cdot xcdot1}-157x^3=left(9sqrt[9]{left(frac{185}{3}right)^6cdot3}-157right)x^3geq0.$$
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add a comment |
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1 Answer
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$begingroup$
Because by AM-GM $$1+x+3x^2-157x^3+415x^4-225x^5+381x^6-99x^7=$$
$$=1+x+3x^2-157x^3+370x^4+x^4(45-225x+282x^2)+99x^6(1-x)geq$$
$$geq370x^4+3x^2+x+1-157x^3=6cdotfrac{185}{3}x^4+3x^2+x+1-157x^3ge$$
$$geq9sqrt[9]{left(frac{185}{3}x^4right)^6cdot3x^2cdot xcdot1}-157x^3=left(9sqrt[9]{left(frac{185}{3}right)^6cdot3}-157right)x^3geq0.$$
$endgroup$
add a comment |
$begingroup$
Because by AM-GM $$1+x+3x^2-157x^3+415x^4-225x^5+381x^6-99x^7=$$
$$=1+x+3x^2-157x^3+370x^4+x^4(45-225x+282x^2)+99x^6(1-x)geq$$
$$geq370x^4+3x^2+x+1-157x^3=6cdotfrac{185}{3}x^4+3x^2+x+1-157x^3ge$$
$$geq9sqrt[9]{left(frac{185}{3}x^4right)^6cdot3x^2cdot xcdot1}-157x^3=left(9sqrt[9]{left(frac{185}{3}right)^6cdot3}-157right)x^3geq0.$$
$endgroup$
add a comment |
$begingroup$
Because by AM-GM $$1+x+3x^2-157x^3+415x^4-225x^5+381x^6-99x^7=$$
$$=1+x+3x^2-157x^3+370x^4+x^4(45-225x+282x^2)+99x^6(1-x)geq$$
$$geq370x^4+3x^2+x+1-157x^3=6cdotfrac{185}{3}x^4+3x^2+x+1-157x^3ge$$
$$geq9sqrt[9]{left(frac{185}{3}x^4right)^6cdot3x^2cdot xcdot1}-157x^3=left(9sqrt[9]{left(frac{185}{3}right)^6cdot3}-157right)x^3geq0.$$
$endgroup$
Because by AM-GM $$1+x+3x^2-157x^3+415x^4-225x^5+381x^6-99x^7=$$
$$=1+x+3x^2-157x^3+370x^4+x^4(45-225x+282x^2)+99x^6(1-x)geq$$
$$geq370x^4+3x^2+x+1-157x^3=6cdotfrac{185}{3}x^4+3x^2+x+1-157x^3ge$$
$$geq9sqrt[9]{left(frac{185}{3}x^4right)^6cdot3x^2cdot xcdot1}-157x^3=left(9sqrt[9]{left(frac{185}{3}right)^6cdot3}-157right)x^3geq0.$$
edited Dec 19 '18 at 13:08
answered Dec 19 '18 at 13:01
Michael RozenbergMichael Rozenberg
109k1896201
109k1896201
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$begingroup$
Not sure this helps, but here's what the part closest to the $x$-axis looks like for $x < 1$: wolframalpha.com/input/…
$endgroup$
– David K
Dec 19 '18 at 12:44
1
$begingroup$
How about using Sturm's Theorem en.wikipedia.org/wiki/Sturm%27s_theorem and show that the polynomial does not have real roots in the interval (0,1]?
$endgroup$
– Johannes Huisman
Dec 19 '18 at 12:59