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Prove $a+2a^2+3a^31$.



The 2019 Stack Overflow Developer Survey Results Are In
Announcing the arrival of Valued Associate #679: Cesar Manara
Planned maintenance scheduled April 17/18, 2019 at 00:00UTC (8:00pm US/Eastern)Basic Proof QuestionCombinatorics question about additionBasic Algebra problem giving me problemsProving some trig identities.Revisiting algebra for the proofsSolving triangles with trig, word problemTeacher ResourceWhy is math so difficult for me?Quadratic equation - What is the value of x?I cannot comprehend ANY math. I cannot understand how things can be equal yet separate.










1












$begingroup$


I feel like this is way easier than my work until now has shown, but I'm stuck and don't even know what to try.



This isn't supposed to be too hard and I know that I'm not supposed to use any special algebraic identities, just basic things. So I tried subtracting terms from the left side from the right side to work backwards, for e.g: $a+2a^2+3a^3<6a^6implies a+3a^3<3a^3(2a^3-1)$, but this has gotten me nowhere. I also noticed that the inequality is $a+2a^2+3a^3<atimes2a^2times3a^3$, but I didn't know what to do with that either.






algebra-precalculus inequality







share|cite|improve this question









$endgroup$







  • 2




    $begingroup$
    Use the fact that for $n<m$ and $a>1$ we have $a^n<a^m$.
    $endgroup$
    – Floris Claassens
    Apr 8 at 12:26















1












$begingroup$


I feel like this is way easier than my work until now has shown, but I'm stuck and don't even know what to try.



This isn't supposed to be too hard and I know that I'm not supposed to use any special algebraic identities, just basic things. So I tried subtracting terms from the left side from the right side to work backwards, for e.g: $a+2a^2+3a^3<6a^6implies a+3a^3<3a^3(2a^3-1)$, but this has gotten me nowhere. I also noticed that the inequality is $a+2a^2+3a^3<atimes2a^2times3a^3$, but I didn't know what to do with that either.






algebra-precalculus inequality







share|cite|improve this question









$endgroup$







  • 2




    $begingroup$
    Use the fact that for $n<m$ and $a>1$ we have $a^n<a^m$.
    $endgroup$
    – Floris Claassens
    Apr 8 at 12:26













1












1








1





$begingroup$


I feel like this is way easier than my work until now has shown, but I'm stuck and don't even know what to try.



This isn't supposed to be too hard and I know that I'm not supposed to use any special algebraic identities, just basic things. So I tried subtracting terms from the left side from the right side to work backwards, for e.g: $a+2a^2+3a^3<6a^6implies a+3a^3<3a^3(2a^3-1)$, but this has gotten me nowhere. I also noticed that the inequality is $a+2a^2+3a^3<atimes2a^2times3a^3$, but I didn't know what to do with that either.






algebra-precalculus inequality







share|cite|improve this question









$endgroup$




I feel like this is way easier than my work until now has shown, but I'm stuck and don't even know what to try.



This isn't supposed to be too hard and I know that I'm not supposed to use any special algebraic identities, just basic things. So I tried subtracting terms from the left side from the right side to work backwards, for e.g: $a+2a^2+3a^3<6a^6implies a+3a^3<3a^3(2a^3-1)$, but this has gotten me nowhere. I also noticed that the inequality is $a+2a^2+3a^3<atimes2a^2times3a^3$, but I didn't know what to do with that either.






algebra-precalculus inequality




algebra-precalculus inequality






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Apr 8 at 12:24









huB1erTi2huB1erTi2

668




668







  • 2




    $begingroup$
    Use the fact that for $n<m$ and $a>1$ we have $a^n<a^m$.
    $endgroup$
    – Floris Claassens
    Apr 8 at 12:26












  • 2




    $begingroup$
    Use the fact that for $n<m$ and $a>1$ we have $a^n<a^m$.
    $endgroup$
    – Floris Claassens
    Apr 8 at 12:26







2




2




$begingroup$
Use the fact that for $n<m$ and $a>1$ we have $a^n<a^m$.
$endgroup$
– Floris Claassens
Apr 8 at 12:26




$begingroup$
Use the fact that for $n<m$ and $a>1$ we have $a^n<a^m$.
$endgroup$
– Floris Claassens
Apr 8 at 12:26










2 Answers
2






active

oldest

votes


















7












$begingroup$

Hint: $a < a^6$, $a^2 < a^6$, $a^3 < a^6$ for $a > 1$.






share|cite|improve this answer











$endgroup$








  • 2




    $begingroup$
    Oh my ******* God. Of course. Love it when I get blocked and think too far. Thanks for the answer.
    $endgroup$
    – huB1erTi2
    Apr 8 at 12:28


















5












$begingroup$

Note that
$$
6a^6 -3a^3-2a^2-a=(6a^4 + 6a^3 + 6a^2 + 3a + 1)(a - 1)a>0
$$
for all $a>1$.






share|cite|improve this answer









$endgroup$












  • $begingroup$
    You complicated it too much (as I did in my attempts), but at least it worked for you, thanks.
    $endgroup$
    – huB1erTi2
    Apr 8 at 12:29






  • 2




    $begingroup$
    All they did was factorise the result by realising the polynomials obvious roots at $0$ and $1$?
    $endgroup$
    – Peter Foreman
    Apr 8 at 12:29











Your Answer








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2 Answers
2






active

oldest

votes








2 Answers
2






active

oldest

votes









active

oldest

votes






active

oldest

votes









7












$begingroup$

Hint: $a < a^6$, $a^2 < a^6$, $a^3 < a^6$ for $a > 1$.






share|cite|improve this answer











$endgroup$








  • 2




    $begingroup$
    Oh my ******* God. Of course. Love it when I get blocked and think too far. Thanks for the answer.
    $endgroup$
    – huB1erTi2
    Apr 8 at 12:28















7












$begingroup$

Hint: $a < a^6$, $a^2 < a^6$, $a^3 < a^6$ for $a > 1$.






share|cite|improve this answer











$endgroup$








  • 2




    $begingroup$
    Oh my ******* God. Of course. Love it when I get blocked and think too far. Thanks for the answer.
    $endgroup$
    – huB1erTi2
    Apr 8 at 12:28













7












7








7





$begingroup$

Hint: $a < a^6$, $a^2 < a^6$, $a^3 < a^6$ for $a > 1$.






share|cite|improve this answer











$endgroup$



Hint: $a < a^6$, $a^2 < a^6$, $a^3 < a^6$ for $a > 1$.







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Apr 8 at 12:49









huB1erTi2

668




668










answered Apr 8 at 12:26









Robert IsraelRobert Israel

331k23221477




331k23221477







  • 2




    $begingroup$
    Oh my ******* God. Of course. Love it when I get blocked and think too far. Thanks for the answer.
    $endgroup$
    – huB1erTi2
    Apr 8 at 12:28












  • 2




    $begingroup$
    Oh my ******* God. Of course. Love it when I get blocked and think too far. Thanks for the answer.
    $endgroup$
    – huB1erTi2
    Apr 8 at 12:28







2




2




$begingroup$
Oh my ******* God. Of course. Love it when I get blocked and think too far. Thanks for the answer.
$endgroup$
– huB1erTi2
Apr 8 at 12:28




$begingroup$
Oh my ******* God. Of course. Love it when I get blocked and think too far. Thanks for the answer.
$endgroup$
– huB1erTi2
Apr 8 at 12:28











5












$begingroup$

Note that
$$
6a^6 -3a^3-2a^2-a=(6a^4 + 6a^3 + 6a^2 + 3a + 1)(a - 1)a>0
$$
for all $a>1$.






share|cite|improve this answer









$endgroup$












  • $begingroup$
    You complicated it too much (as I did in my attempts), but at least it worked for you, thanks.
    $endgroup$
    – huB1erTi2
    Apr 8 at 12:29






  • 2




    $begingroup$
    All they did was factorise the result by realising the polynomials obvious roots at $0$ and $1$?
    $endgroup$
    – Peter Foreman
    Apr 8 at 12:29















5












$begingroup$

Note that
$$
6a^6 -3a^3-2a^2-a=(6a^4 + 6a^3 + 6a^2 + 3a + 1)(a - 1)a>0
$$
for all $a>1$.






share|cite|improve this answer









$endgroup$












  • $begingroup$
    You complicated it too much (as I did in my attempts), but at least it worked for you, thanks.
    $endgroup$
    – huB1erTi2
    Apr 8 at 12:29






  • 2




    $begingroup$
    All they did was factorise the result by realising the polynomials obvious roots at $0$ and $1$?
    $endgroup$
    – Peter Foreman
    Apr 8 at 12:29













5












5








5





$begingroup$

Note that
$$
6a^6 -3a^3-2a^2-a=(6a^4 + 6a^3 + 6a^2 + 3a + 1)(a - 1)a>0
$$
for all $a>1$.






share|cite|improve this answer









$endgroup$



Note that
$$
6a^6 -3a^3-2a^2-a=(6a^4 + 6a^3 + 6a^2 + 3a + 1)(a - 1)a>0
$$
for all $a>1$.







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Apr 8 at 12:26









Dietrich BurdeDietrich Burde

82k649107




82k649107











  • $begingroup$
    You complicated it too much (as I did in my attempts), but at least it worked for you, thanks.
    $endgroup$
    – huB1erTi2
    Apr 8 at 12:29






  • 2




    $begingroup$
    All they did was factorise the result by realising the polynomials obvious roots at $0$ and $1$?
    $endgroup$
    – Peter Foreman
    Apr 8 at 12:29
















  • $begingroup$
    You complicated it too much (as I did in my attempts), but at least it worked for you, thanks.
    $endgroup$
    – huB1erTi2
    Apr 8 at 12:29






  • 2




    $begingroup$
    All they did was factorise the result by realising the polynomials obvious roots at $0$ and $1$?
    $endgroup$
    – Peter Foreman
    Apr 8 at 12:29















$begingroup$
You complicated it too much (as I did in my attempts), but at least it worked for you, thanks.
$endgroup$
– huB1erTi2
Apr 8 at 12:29




$begingroup$
You complicated it too much (as I did in my attempts), but at least it worked for you, thanks.
$endgroup$
– huB1erTi2
Apr 8 at 12:29




2




2




$begingroup$
All they did was factorise the result by realising the polynomials obvious roots at $0$ and $1$?
$endgroup$
– Peter Foreman
Apr 8 at 12:29




$begingroup$
All they did was factorise the result by realising the polynomials obvious roots at $0$ and $1$?
$endgroup$
– Peter Foreman
Apr 8 at 12:29

















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