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.
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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.
$begingroup$
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
asked Apr 8 at 12:24
huB1erTi2huB1erTi2
668
$endgroup$
add a comment |
$begingroup$
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
asked Apr 8 at 12:24
huB1erTi2huB1erTi2
668
$endgroup$
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
add a comment |
$begingroup$
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
asked Apr 8 at 12:24
huB1erTi2huB1erTi2
668
$endgroup$
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
asked Apr 8 at 12:24
huB1erTi2huB1erTi2
668
asked Apr 8 at 12:24
huB1erTi2huB1erTi2
668
asked Apr 8 at 12:24
huB1erTi2huB1erTi2
668
asked Apr 8 at 12:24
huB1erTi2huB1erTi2
668
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
add a comment |
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
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
Hint: $a < a^6$, $a^2 < a^6$, $a^3 < a^6$ for $a > 1$.
edited Apr 8 at 12:49
huB1erTi2
668
answered Apr 8 at 12:26
Robert IsraelRobert Israel
331k23221477
$endgroup$
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
add a comment |
$begingroup$
Note that
$$
6a^6 -3a^3-2a^2-a=(6a^4 + 6a^3 + 6a^2 + 3a + 1)(a - 1)a>0
$$
for all $a>1$.
answered Apr 8 at 12:26
Dietrich BurdeDietrich Burde
82k649107
$endgroup$
$begingroup$
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
add a comment |
Your Answer
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Hint: $a < a^6$, $a^2 < a^6$, $a^3 < a^6$ for $a > 1$.
edited Apr 8 at 12:49
huB1erTi2
668
answered Apr 8 at 12:26
Robert IsraelRobert Israel
331k23221477
$endgroup$
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
add a comment |
$begingroup$
Hint: $a < a^6$, $a^2 < a^6$, $a^3 < a^6$ for $a > 1$.
edited Apr 8 at 12:49
huB1erTi2
668
answered Apr 8 at 12:26
Robert IsraelRobert Israel
331k23221477
$endgroup$
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
add a comment |
$begingroup$
Hint: $a < a^6$, $a^2 < a^6$, $a^3 < a^6$ for $a > 1$.
edited Apr 8 at 12:49
huB1erTi2
668
answered Apr 8 at 12:26
Robert IsraelRobert Israel
331k23221477
$endgroup$
Hint: $a < a^6$, $a^2 < a^6$, $a^3 < a^6$ for $a > 1$.
edited Apr 8 at 12:49
huB1erTi2
668
answered Apr 8 at 12:26
Robert IsraelRobert Israel
331k23221477
edited Apr 8 at 12:49
huB1erTi2
668
edited Apr 8 at 12:49
huB1erTi2
668
edited Apr 8 at 12:49
huB1erTi2
668
668
answered Apr 8 at 12:26
Robert IsraelRobert Israel
331k23221477
answered Apr 8 at 12:26
Robert IsraelRobert Israel
331k23221477
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
add a comment |
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
add a comment |
$begingroup$
Note that
$$
6a^6 -3a^3-2a^2-a=(6a^4 + 6a^3 + 6a^2 + 3a + 1)(a - 1)a>0
$$
for all $a>1$.
answered Apr 8 at 12:26
Dietrich BurdeDietrich Burde
82k649107
$endgroup$
$begingroup$
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
add a comment |
$begingroup$
Note that
$$
6a^6 -3a^3-2a^2-a=(6a^4 + 6a^3 + 6a^2 + 3a + 1)(a - 1)a>0
$$
for all $a>1$.
answered Apr 8 at 12:26
Dietrich BurdeDietrich Burde
82k649107
$endgroup$
$begingroup$
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
add a comment |
$begingroup$
Note that
$$
6a^6 -3a^3-2a^2-a=(6a^4 + 6a^3 + 6a^2 + 3a + 1)(a - 1)a>0
$$
for all $a>1$.
answered Apr 8 at 12:26
Dietrich BurdeDietrich Burde
82k649107
$endgroup$
Note that
$$
6a^6 -3a^3-2a^2-a=(6a^4 + 6a^3 + 6a^2 + 3a + 1)(a - 1)a>0
$$
for all $a>1$.
answered Apr 8 at 12:26
Dietrich BurdeDietrich Burde
82k649107
answered Apr 8 at 12:26
Dietrich BurdeDietrich Burde
82k649107
answered Apr 8 at 12:26
Dietrich BurdeDietrich Burde
82k649107
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
add a comment |
$begingroup$
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
add a comment |
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$begingroup$
Use the fact that for $n<m$ and $a>1$ we have $a^n<a^m$.
$endgroup$
– Floris Claassens
Apr 8 at 12:26