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Proving that if x^4 + 5x + 1 < 27 then x < 2
Clash Royale CLAN TAG#URR8PPP
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2
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I need help trying to figure out how I can prove a statement like this. So looking at this I can conclude that this statement is of the form P -> Q
P : x^4 + 5x + 1 < 27
Q : x < 2
I wanted to try and prove this by contrapositive , so this state would become
If X>=2 then x^4 + 5x + 1 >= 27
Over here I was not sure if I could plug in the value 2 or anythin greater than 2 to see if this is true. plugging in 2 gets me 2^4 + 5*2 + 1 = 27 and since 27>=27 this statement is True.
Am I allowed to prove it like this? Is there a different way to prove a question like this? Any help would be appreciated thank you.
proof-verification proof-writing proof-explanation
asked 1 hour ago
Rufyi
355
add a comment |Â
up vote
2
down vote
favorite
I need help trying to figure out how I can prove a statement like this. So looking at this I can conclude that this statement is of the form P -> Q
P : x^4 + 5x + 1 < 27
Q : x < 2
I wanted to try and prove this by contrapositive , so this state would become
If X>=2 then x^4 + 5x + 1 >= 27
Over here I was not sure if I could plug in the value 2 or anythin greater than 2 to see if this is true. plugging in 2 gets me 2^4 + 5*2 + 1 = 27 and since 27>=27 this statement is True.
Am I allowed to prove it like this? Is there a different way to prove a question like this? Any help would be appreciated thank you.
proof-verification proof-writing proof-explanation
asked 1 hour ago
Rufyi
355
No, you have to prove for a general $xgeq 2$. It might work for a specific value but not in general. That is the problem with your approach
– Heisenberg
1 hour ago
If $x ge 2$ then $x^4 + 5x +1 ge 16+10+1 = 27$.
– copper.hat
1 hour ago
You can use that for $cge0$, $age b$ implies $acge bc$. So for $xge 2$ (and thereby $x>0$) we have $x^2ge 2xge 4$, $x^3ge 2x^2ge 8$, $x^4ge 2x^3ge 16$, and hence by addimg the inequality $x^4+4x+1ge 16+10+1=27$ indeed
– Hagen von Eitzen
1 hour ago
add a comment |Â
up vote
2
down vote
favorite
up vote
2
down vote
favorite
I need help trying to figure out how I can prove a statement like this. So looking at this I can conclude that this statement is of the form P -> Q
P : x^4 + 5x + 1 < 27
Q : x < 2
I wanted to try and prove this by contrapositive , so this state would become
If X>=2 then x^4 + 5x + 1 >= 27
Over here I was not sure if I could plug in the value 2 or anythin greater than 2 to see if this is true. plugging in 2 gets me 2^4 + 5*2 + 1 = 27 and since 27>=27 this statement is True.
Am I allowed to prove it like this? Is there a different way to prove a question like this? Any help would be appreciated thank you.
proof-verification proof-writing proof-explanation
asked 1 hour ago
Rufyi
355
I need help trying to figure out how I can prove a statement like this. So looking at this I can conclude that this statement is of the form P -> Q
P : x^4 + 5x + 1 < 27
Q : x < 2
I wanted to try and prove this by contrapositive , so this state would become
If X>=2 then x^4 + 5x + 1 >= 27
Over here I was not sure if I could plug in the value 2 or anythin greater than 2 to see if this is true. plugging in 2 gets me 2^4 + 5*2 + 1 = 27 and since 27>=27 this statement is True.
Am I allowed to prove it like this? Is there a different way to prove a question like this? Any help would be appreciated thank you.
proof-verification proof-writing proof-explanation
proof-verification proof-writing proof-explanation
asked 1 hour ago
Rufyi
355
asked 1 hour ago
Rufyi
355
asked 1 hour ago
Rufyi
355
asked 1 hour ago
Rufyi
355
asked 1 hour ago
Rufyi
355
355
No, you have to prove for a general $xgeq 2$. It might work for a specific value but not in general. That is the problem with your approach
– Heisenberg
1 hour ago
If $x ge 2$ then $x^4 + 5x +1 ge 16+10+1 = 27$.
– copper.hat
1 hour ago
You can use that for $cge0$, $age b$ implies $acge bc$. So for $xge 2$ (and thereby $x>0$) we have $x^2ge 2xge 4$, $x^3ge 2x^2ge 8$, $x^4ge 2x^3ge 16$, and hence by addimg the inequality $x^4+4x+1ge 16+10+1=27$ indeed
– Hagen von Eitzen
1 hour ago
add a comment |Â
No, you have to prove for a general $xgeq 2$. It might work for a specific value but not in general. That is the problem with your approach
– Heisenberg
1 hour ago
If $x ge 2$ then $x^4 + 5x +1 ge 16+10+1 = 27$.
– copper.hat
1 hour ago
You can use that for $cge0$, $age b$ implies $acge bc$. So for $xge 2$ (and thereby $x>0$) we have $x^2ge 2xge 4$, $x^3ge 2x^2ge 8$, $x^4ge 2x^3ge 16$, and hence by addimg the inequality $x^4+4x+1ge 16+10+1=27$ indeed
– Hagen von Eitzen
1 hour ago
No, you have to prove for a general $xgeq 2$. It might work for a specific value but not in general. That is the problem with your approach
– Heisenberg
1 hour ago
No, you have to prove for a general $xgeq 2$. It might work for a specific value but not in general. That is the problem with your approach
– Heisenberg
1 hour ago
If $x ge 2$ then $x^4 + 5x +1 ge 16+10+1 = 27$.
– copper.hat
1 hour ago
If $x ge 2$ then $x^4 + 5x +1 ge 16+10+1 = 27$.
– copper.hat
1 hour ago
You can use that for $cge0$, $age b$ implies $acge bc$. So for $xge 2$ (and thereby $x>0$) we have $x^2ge 2xge 4$, $x^3ge 2x^2ge 8$, $x^4ge 2x^3ge 16$, and hence by addimg the inequality $x^4+4x+1ge 16+10+1=27$ indeed
– Hagen von Eitzen
1 hour ago
You can use that for $cge0$, $age b$ implies $acge bc$. So for $xge 2$ (and thereby $x>0$) we have $x^2ge 2xge 4$, $x^3ge 2x^2ge 8$, $x^4ge 2x^3ge 16$, and hence by addimg the inequality $x^4+4x+1ge 16+10+1=27$ indeed
– Hagen von Eitzen
1 hour ago
add a comment |Â
4 Answers
4
active
oldest
votes
up vote
3
down vote
accepted
Suppose $xgeq2$. Then $x^4geq16implies x^4
+5xgeq16+5xgeq 16+...$. Got the hint?
answered 1 hour ago
Heisenberg
1,3911639
Are you trying to make the left side look like what we need to prove?
– Rufyi
1 hour ago
If I start from x^4 >= 16, I would get x^4 + 5x + 1 >= 16 + 5x + 1, can i plug in 2 in for 5x on the right side?
– Rufyi
1 hour ago
Yes exactly!...
– Heisenberg
1 hour ago
Can I do what I said in the second comment?
– Rufyi
58 mins ago
Yes when you plug in 2, you get 27 on the right side thus proving $x^4+5x+1geq 27$
– Heisenberg
56 mins ago
add a comment |Â
up vote
1
down vote
A marginally different take:
Let $p(x) = x^4+5x-26$, we would like to show that if $p(x)<0$ then $x <2$.
Note that $p(2) = 0$ and so synthetic division gives $p(x) = (x-2)(x^3+2x^2+4x+13)$.
In particular, note that $x^3+2x^2+4x+13$ has no positive roots. Hence if $x ge 0$,
we see that $p(x)$ and $x-2$ have the same sign, hence if $p(x) < 0$ we must have
$x < 2$.
answered 1 hour ago
copper.hat
124k558156
add a comment |Â
up vote
0
down vote
Trying to show the contrapositive is a valid method, but you aren't allowed to just give an example. For example, if you wanted to prove that
$$xgeq 2implies x^4+5x+1leq 27,$$
which is false for example at $x=0$, you can't just give the example of $x=2$.
One hint for trying to prove this is to show that $f(x)=x^4+5x+1$ is increasing over the positive real numbers - in other words, assume that $0<a<b$ and show that $f(b)>f(a)$.
answered 1 hour ago
Carl Schildkraut
9,82911238
add a comment |Â
up vote
0
down vote
If $xge 2$ is it true that $x^4 ge 16$?
If $x ge 2$ is it true that $5x ge 10$?
So if $x ge 2$ is it true that $x^4 + 5x + 1 ge 16 + 10 + 1 = 27$?
...
It could get tedious to go to axioms and prove that if $xge 2 > 0$ then $x^4 ge 2x^3 ge 4x^2 ge 8x ge 16$ via the axiom: if $a > 0$ and $m < n$ then $am < an$ (applied three times); and $5 > 0$ so $xge 2 implies 5x ge 2*5 =10$ by the same axiom; and by the axiom $a + c > b + c iff a > b$ then $x^4 + 5x + 1 ge x^4 + 10 + 1 ge 16+ 10 + 1=27$. But I think it is safe to assume if we had to prove those we could.
This may be tedious over kill but:
If $x ge 2$ then let $d = x - 2 ge 0$. Then $x = 2 + d$
Now because $d ge 0$ then $d^k ge 0$.
So $x^4 = (2 + d)^4 = 2^4 + 4*2^3*d + 6*2^2*d^2 + 4*2*d^3 + d^4 ge 2^4$
And $5x = 5*(2+d) = 5*2 + 5*d ge 5*2$
So $x^4 + 5x + 1ge 2^4 + 5*2 + 1 = 27$.
That should do it.....
answered 46 mins ago
fleablood
63.4k22679
add a comment |Â
4 Answers
4
active
oldest
votes
4 Answers
4
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
3
down vote
accepted
Suppose $xgeq2$. Then $x^4geq16implies x^4
+5xgeq16+5xgeq 16+...$. Got the hint?
answered 1 hour ago
Heisenberg
1,3911639
Are you trying to make the left side look like what we need to prove?
– Rufyi
1 hour ago
If I start from x^4 >= 16, I would get x^4 + 5x + 1 >= 16 + 5x + 1, can i plug in 2 in for 5x on the right side?
– Rufyi
1 hour ago
Yes exactly!...
– Heisenberg
1 hour ago
Can I do what I said in the second comment?
– Rufyi
58 mins ago
Yes when you plug in 2, you get 27 on the right side thus proving $x^4+5x+1geq 27$
– Heisenberg
56 mins ago
add a comment |Â
up vote
3
down vote
accepted
Suppose $xgeq2$. Then $x^4geq16implies x^4
+5xgeq16+5xgeq 16+...$. Got the hint?
answered 1 hour ago
Heisenberg
1,3911639
Are you trying to make the left side look like what we need to prove?
– Rufyi
1 hour ago
If I start from x^4 >= 16, I would get x^4 + 5x + 1 >= 16 + 5x + 1, can i plug in 2 in for 5x on the right side?
– Rufyi
1 hour ago
Yes exactly!...
– Heisenberg
1 hour ago
Can I do what I said in the second comment?
– Rufyi
58 mins ago
Yes when you plug in 2, you get 27 on the right side thus proving $x^4+5x+1geq 27$
– Heisenberg
56 mins ago
add a comment |Â
up vote
3
down vote
accepted
up vote
3
down vote
accepted
Suppose $xgeq2$. Then $x^4geq16implies x^4
+5xgeq16+5xgeq 16+...$. Got the hint?
answered 1 hour ago
Heisenberg
1,3911639
Suppose $xgeq2$. Then $x^4geq16implies x^4
+5xgeq16+5xgeq 16+...$. Got the hint?
answered 1 hour ago
Heisenberg
1,3911639
answered 1 hour ago
Heisenberg
1,3911639
answered 1 hour ago
Heisenberg
1,3911639
answered 1 hour ago
Heisenberg
1,3911639
1,3911639
Are you trying to make the left side look like what we need to prove?
– Rufyi
1 hour ago
If I start from x^4 >= 16, I would get x^4 + 5x + 1 >= 16 + 5x + 1, can i plug in 2 in for 5x on the right side?
– Rufyi
1 hour ago
Yes exactly!...
– Heisenberg
1 hour ago
Can I do what I said in the second comment?
– Rufyi
58 mins ago
Yes when you plug in 2, you get 27 on the right side thus proving $x^4+5x+1geq 27$
– Heisenberg
56 mins ago
add a comment |Â
Are you trying to make the left side look like what we need to prove?
– Rufyi
1 hour ago
If I start from x^4 >= 16, I would get x^4 + 5x + 1 >= 16 + 5x + 1, can i plug in 2 in for 5x on the right side?
– Rufyi
1 hour ago
Yes exactly!...
– Heisenberg
1 hour ago
Can I do what I said in the second comment?
– Rufyi
58 mins ago
Yes when you plug in 2, you get 27 on the right side thus proving $x^4+5x+1geq 27$
– Heisenberg
56 mins ago
Are you trying to make the left side look like what we need to prove?
– Rufyi
1 hour ago
Are you trying to make the left side look like what we need to prove?
– Rufyi
1 hour ago
If I start from x^4 >= 16, I would get x^4 + 5x + 1 >= 16 + 5x + 1, can i plug in 2 in for 5x on the right side?
– Rufyi
1 hour ago
If I start from x^4 >= 16, I would get x^4 + 5x + 1 >= 16 + 5x + 1, can i plug in 2 in for 5x on the right side?
– Rufyi
1 hour ago
Yes exactly!...
– Heisenberg
1 hour ago
Yes exactly!...
– Heisenberg
1 hour ago
Can I do what I said in the second comment?
– Rufyi
58 mins ago
Can I do what I said in the second comment?
– Rufyi
58 mins ago
Yes when you plug in 2, you get 27 on the right side thus proving $x^4+5x+1geq 27$
– Heisenberg
56 mins ago
Yes when you plug in 2, you get 27 on the right side thus proving $x^4+5x+1geq 27$
– Heisenberg
56 mins ago
add a comment |Â
up vote
1
down vote
A marginally different take:
Let $p(x) = x^4+5x-26$, we would like to show that if $p(x)<0$ then $x <2$.
Note that $p(2) = 0$ and so synthetic division gives $p(x) = (x-2)(x^3+2x^2+4x+13)$.
In particular, note that $x^3+2x^2+4x+13$ has no positive roots. Hence if $x ge 0$,
we see that $p(x)$ and $x-2$ have the same sign, hence if $p(x) < 0$ we must have
$x < 2$.
answered 1 hour ago
copper.hat
124k558156
add a comment |Â
up vote
1
down vote
A marginally different take:
Let $p(x) = x^4+5x-26$, we would like to show that if $p(x)<0$ then $x <2$.
Note that $p(2) = 0$ and so synthetic division gives $p(x) = (x-2)(x^3+2x^2+4x+13)$.
In particular, note that $x^3+2x^2+4x+13$ has no positive roots. Hence if $x ge 0$,
we see that $p(x)$ and $x-2$ have the same sign, hence if $p(x) < 0$ we must have
$x < 2$.
answered 1 hour ago
copper.hat
124k558156
add a comment |Â
up vote
1
down vote
up vote
1
down vote
A marginally different take:
Let $p(x) = x^4+5x-26$, we would like to show that if $p(x)<0$ then $x <2$.
Note that $p(2) = 0$ and so synthetic division gives $p(x) = (x-2)(x^3+2x^2+4x+13)$.
In particular, note that $x^3+2x^2+4x+13$ has no positive roots. Hence if $x ge 0$,
we see that $p(x)$ and $x-2$ have the same sign, hence if $p(x) < 0$ we must have
$x < 2$.
answered 1 hour ago
copper.hat
124k558156
A marginally different take:
Let $p(x) = x^4+5x-26$, we would like to show that if $p(x)<0$ then $x <2$.
Note that $p(2) = 0$ and so synthetic division gives $p(x) = (x-2)(x^3+2x^2+4x+13)$.
In particular, note that $x^3+2x^2+4x+13$ has no positive roots. Hence if $x ge 0$,
we see that $p(x)$ and $x-2$ have the same sign, hence if $p(x) < 0$ we must have
$x < 2$.
answered 1 hour ago
copper.hat
124k558156
answered 1 hour ago
copper.hat
124k558156
answered 1 hour ago
copper.hat
124k558156
answered 1 hour ago
copper.hat
124k558156
124k558156
add a comment |Â
add a comment |Â
up vote
0
down vote
Trying to show the contrapositive is a valid method, but you aren't allowed to just give an example. For example, if you wanted to prove that
$$xgeq 2implies x^4+5x+1leq 27,$$
which is false for example at $x=0$, you can't just give the example of $x=2$.
One hint for trying to prove this is to show that $f(x)=x^4+5x+1$ is increasing over the positive real numbers - in other words, assume that $0<a<b$ and show that $f(b)>f(a)$.
answered 1 hour ago
Carl Schildkraut
9,82911238
add a comment |Â
up vote
0
down vote
Trying to show the contrapositive is a valid method, but you aren't allowed to just give an example. For example, if you wanted to prove that
$$xgeq 2implies x^4+5x+1leq 27,$$
which is false for example at $x=0$, you can't just give the example of $x=2$.
One hint for trying to prove this is to show that $f(x)=x^4+5x+1$ is increasing over the positive real numbers - in other words, assume that $0<a<b$ and show that $f(b)>f(a)$.
answered 1 hour ago
Carl Schildkraut
9,82911238
add a comment |Â
up vote
0
down vote
up vote
0
down vote
Trying to show the contrapositive is a valid method, but you aren't allowed to just give an example. For example, if you wanted to prove that
$$xgeq 2implies x^4+5x+1leq 27,$$
which is false for example at $x=0$, you can't just give the example of $x=2$.
One hint for trying to prove this is to show that $f(x)=x^4+5x+1$ is increasing over the positive real numbers - in other words, assume that $0<a<b$ and show that $f(b)>f(a)$.
answered 1 hour ago
Carl Schildkraut
9,82911238
Trying to show the contrapositive is a valid method, but you aren't allowed to just give an example. For example, if you wanted to prove that
$$xgeq 2implies x^4+5x+1leq 27,$$
which is false for example at $x=0$, you can't just give the example of $x=2$.
One hint for trying to prove this is to show that $f(x)=x^4+5x+1$ is increasing over the positive real numbers - in other words, assume that $0<a<b$ and show that $f(b)>f(a)$.
answered 1 hour ago
Carl Schildkraut
9,82911238
answered 1 hour ago
Carl Schildkraut
9,82911238
answered 1 hour ago
Carl Schildkraut
9,82911238
answered 1 hour ago
Carl Schildkraut
9,82911238
9,82911238
add a comment |Â
add a comment |Â
up vote
0
down vote
If $xge 2$ is it true that $x^4 ge 16$?
If $x ge 2$ is it true that $5x ge 10$?
So if $x ge 2$ is it true that $x^4 + 5x + 1 ge 16 + 10 + 1 = 27$?
...
It could get tedious to go to axioms and prove that if $xge 2 > 0$ then $x^4 ge 2x^3 ge 4x^2 ge 8x ge 16$ via the axiom: if $a > 0$ and $m < n$ then $am < an$ (applied three times); and $5 > 0$ so $xge 2 implies 5x ge 2*5 =10$ by the same axiom; and by the axiom $a + c > b + c iff a > b$ then $x^4 + 5x + 1 ge x^4 + 10 + 1 ge 16+ 10 + 1=27$. But I think it is safe to assume if we had to prove those we could.
This may be tedious over kill but:
If $x ge 2$ then let $d = x - 2 ge 0$. Then $x = 2 + d$
Now because $d ge 0$ then $d^k ge 0$.
So $x^4 = (2 + d)^4 = 2^4 + 4*2^3*d + 6*2^2*d^2 + 4*2*d^3 + d^4 ge 2^4$
And $5x = 5*(2+d) = 5*2 + 5*d ge 5*2$
So $x^4 + 5x + 1ge 2^4 + 5*2 + 1 = 27$.
That should do it.....
answered 46 mins ago
fleablood
63.4k22679
add a comment |Â
up vote
0
down vote
If $xge 2$ is it true that $x^4 ge 16$?
If $x ge 2$ is it true that $5x ge 10$?
So if $x ge 2$ is it true that $x^4 + 5x + 1 ge 16 + 10 + 1 = 27$?
...
It could get tedious to go to axioms and prove that if $xge 2 > 0$ then $x^4 ge 2x^3 ge 4x^2 ge 8x ge 16$ via the axiom: if $a > 0$ and $m < n$ then $am < an$ (applied three times); and $5 > 0$ so $xge 2 implies 5x ge 2*5 =10$ by the same axiom; and by the axiom $a + c > b + c iff a > b$ then $x^4 + 5x + 1 ge x^4 + 10 + 1 ge 16+ 10 + 1=27$. But I think it is safe to assume if we had to prove those we could.
This may be tedious over kill but:
If $x ge 2$ then let $d = x - 2 ge 0$. Then $x = 2 + d$
Now because $d ge 0$ then $d^k ge 0$.
So $x^4 = (2 + d)^4 = 2^4 + 4*2^3*d + 6*2^2*d^2 + 4*2*d^3 + d^4 ge 2^4$
And $5x = 5*(2+d) = 5*2 + 5*d ge 5*2$
So $x^4 + 5x + 1ge 2^4 + 5*2 + 1 = 27$.
That should do it.....
answered 46 mins ago
fleablood
63.4k22679
add a comment |Â
up vote
0
down vote
up vote
0
down vote
If $xge 2$ is it true that $x^4 ge 16$?
If $x ge 2$ is it true that $5x ge 10$?
So if $x ge 2$ is it true that $x^4 + 5x + 1 ge 16 + 10 + 1 = 27$?
...
It could get tedious to go to axioms and prove that if $xge 2 > 0$ then $x^4 ge 2x^3 ge 4x^2 ge 8x ge 16$ via the axiom: if $a > 0$ and $m < n$ then $am < an$ (applied three times); and $5 > 0$ so $xge 2 implies 5x ge 2*5 =10$ by the same axiom; and by the axiom $a + c > b + c iff a > b$ then $x^4 + 5x + 1 ge x^4 + 10 + 1 ge 16+ 10 + 1=27$. But I think it is safe to assume if we had to prove those we could.
This may be tedious over kill but:
If $x ge 2$ then let $d = x - 2 ge 0$. Then $x = 2 + d$
Now because $d ge 0$ then $d^k ge 0$.
So $x^4 = (2 + d)^4 = 2^4 + 4*2^3*d + 6*2^2*d^2 + 4*2*d^3 + d^4 ge 2^4$
And $5x = 5*(2+d) = 5*2 + 5*d ge 5*2$
So $x^4 + 5x + 1ge 2^4 + 5*2 + 1 = 27$.
That should do it.....
answered 46 mins ago
fleablood
63.4k22679
If $xge 2$ is it true that $x^4 ge 16$?
If $x ge 2$ is it true that $5x ge 10$?
So if $x ge 2$ is it true that $x^4 + 5x + 1 ge 16 + 10 + 1 = 27$?
...
It could get tedious to go to axioms and prove that if $xge 2 > 0$ then $x^4 ge 2x^3 ge 4x^2 ge 8x ge 16$ via the axiom: if $a > 0$ and $m < n$ then $am < an$ (applied three times); and $5 > 0$ so $xge 2 implies 5x ge 2*5 =10$ by the same axiom; and by the axiom $a + c > b + c iff a > b$ then $x^4 + 5x + 1 ge x^4 + 10 + 1 ge 16+ 10 + 1=27$. But I think it is safe to assume if we had to prove those we could.
This may be tedious over kill but:
If $x ge 2$ then let $d = x - 2 ge 0$. Then $x = 2 + d$
Now because $d ge 0$ then $d^k ge 0$.
So $x^4 = (2 + d)^4 = 2^4 + 4*2^3*d + 6*2^2*d^2 + 4*2*d^3 + d^4 ge 2^4$
And $5x = 5*(2+d) = 5*2 + 5*d ge 5*2$
So $x^4 + 5x + 1ge 2^4 + 5*2 + 1 = 27$.
That should do it.....
answered 46 mins ago
fleablood
63.4k22679
edited 34 mins ago
answered 46 mins ago
fleablood
63.4k22679
answered 46 mins ago
fleablood
63.4k22679
answered 46 mins ago
fleablood
63.4k22679
63.4k22679
add a comment |Â
add a comment |Â
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No, you have to prove for a general $xgeq 2$. It might work for a specific value but not in general. That is the problem with your approach
– Heisenberg
1 hour ago
If $x ge 2$ then $x^4 + 5x +1 ge 16+10+1 = 27$.
– copper.hat
1 hour ago
You can use that for $cge0$, $age b$ implies $acge bc$. So for $xge 2$ (and thereby $x>0$) we have $x^2ge 2xge 4$, $x^3ge 2x^2ge 8$, $x^4ge 2x^3ge 16$, and hence by addimg the inequality $x^4+4x+1ge 16+10+1=27$ indeed
– Hagen von Eitzen
1 hour ago