Prove that $a+b+cle 0$
Clash Royale CLAN TAG#URR8PPP
up vote
2
down vote
favorite
$a,b,c$ are integer numbers different from zero and
$$fracab+c^2=fraca+c^2b$$
Prove that $a+b+cle 0$.
I know how to prove that $a+b<0$ but do not know what about $c$.
linear-algebra algebra-precalculus inequality
add a comment |Â
up vote
2
down vote
favorite
$a,b,c$ are integer numbers different from zero and
$$fracab+c^2=fraca+c^2b$$
Prove that $a+b+cle 0$.
I know how to prove that $a+b<0$ but do not know what about $c$.
linear-algebra algebra-precalculus inequality
add a comment |Â
up vote
2
down vote
favorite
up vote
2
down vote
favorite
$a,b,c$ are integer numbers different from zero and
$$fracab+c^2=fraca+c^2b$$
Prove that $a+b+cle 0$.
I know how to prove that $a+b<0$ but do not know what about $c$.
linear-algebra algebra-precalculus inequality
$a,b,c$ are integer numbers different from zero and
$$fracab+c^2=fraca+c^2b$$
Prove that $a+b+cle 0$.
I know how to prove that $a+b<0$ but do not know what about $c$.
linear-algebra algebra-precalculus inequality
linear-algebra algebra-precalculus inequality
edited Sep 1 at 9:19
greedoid
28k93776
28k93776
asked Sep 1 at 6:54
math.trouble
335
335
add a comment |Â
add a comment |Â
3 Answers
3
active
oldest
votes
up vote
4
down vote
accepted
Since $c in mathbbZ$, we have $c le |c|le c^2$
$$0=(a+b)c^2+c^4$$
$$a+b+c^2=0$$
$$a+b+c le a+b+c^2=0$$
add a comment |Â
up vote
2
down vote
We have $$ab=(a+c^2)(b+c^2)\ab=ab+c^4+c^2(a+b)$$which means that $$a+b=-c^2le -c$$or equivalently $$a+b+cle 0$$
add a comment |Â
up vote
2
down vote
From given equation we get $a+b+c^2 =0$ (since $cne 0$), so we have $$a+b+c= c-c^2 = c(1-c) leq 0$$
If $c>0$ then $1-cleq 0$ so $c(1-c)leq 0$ and
if $c<0$ then $1-c>0$ so $c(1-c)<0$
add a comment |Â
3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
4
down vote
accepted
Since $c in mathbbZ$, we have $c le |c|le c^2$
$$0=(a+b)c^2+c^4$$
$$a+b+c^2=0$$
$$a+b+c le a+b+c^2=0$$
add a comment |Â
up vote
4
down vote
accepted
Since $c in mathbbZ$, we have $c le |c|le c^2$
$$0=(a+b)c^2+c^4$$
$$a+b+c^2=0$$
$$a+b+c le a+b+c^2=0$$
add a comment |Â
up vote
4
down vote
accepted
up vote
4
down vote
accepted
Since $c in mathbbZ$, we have $c le |c|le c^2$
$$0=(a+b)c^2+c^4$$
$$a+b+c^2=0$$
$$a+b+c le a+b+c^2=0$$
Since $c in mathbbZ$, we have $c le |c|le c^2$
$$0=(a+b)c^2+c^4$$
$$a+b+c^2=0$$
$$a+b+c le a+b+c^2=0$$
answered Sep 1 at 7:01
Siong Thye Goh
81.8k1456104
81.8k1456104
add a comment |Â
add a comment |Â
up vote
2
down vote
We have $$ab=(a+c^2)(b+c^2)\ab=ab+c^4+c^2(a+b)$$which means that $$a+b=-c^2le -c$$or equivalently $$a+b+cle 0$$
add a comment |Â
up vote
2
down vote
We have $$ab=(a+c^2)(b+c^2)\ab=ab+c^4+c^2(a+b)$$which means that $$a+b=-c^2le -c$$or equivalently $$a+b+cle 0$$
add a comment |Â
up vote
2
down vote
up vote
2
down vote
We have $$ab=(a+c^2)(b+c^2)\ab=ab+c^4+c^2(a+b)$$which means that $$a+b=-c^2le -c$$or equivalently $$a+b+cle 0$$
We have $$ab=(a+c^2)(b+c^2)\ab=ab+c^4+c^2(a+b)$$which means that $$a+b=-c^2le -c$$or equivalently $$a+b+cle 0$$
answered Sep 1 at 7:00
Mostafa Ayaz
10.3k3730
10.3k3730
add a comment |Â
add a comment |Â
up vote
2
down vote
From given equation we get $a+b+c^2 =0$ (since $cne 0$), so we have $$a+b+c= c-c^2 = c(1-c) leq 0$$
If $c>0$ then $1-cleq 0$ so $c(1-c)leq 0$ and
if $c<0$ then $1-c>0$ so $c(1-c)<0$
add a comment |Â
up vote
2
down vote
From given equation we get $a+b+c^2 =0$ (since $cne 0$), so we have $$a+b+c= c-c^2 = c(1-c) leq 0$$
If $c>0$ then $1-cleq 0$ so $c(1-c)leq 0$ and
if $c<0$ then $1-c>0$ so $c(1-c)<0$
add a comment |Â
up vote
2
down vote
up vote
2
down vote
From given equation we get $a+b+c^2 =0$ (since $cne 0$), so we have $$a+b+c= c-c^2 = c(1-c) leq 0$$
If $c>0$ then $1-cleq 0$ so $c(1-c)leq 0$ and
if $c<0$ then $1-c>0$ so $c(1-c)<0$
From given equation we get $a+b+c^2 =0$ (since $cne 0$), so we have $$a+b+c= c-c^2 = c(1-c) leq 0$$
If $c>0$ then $1-cleq 0$ so $c(1-c)leq 0$ and
if $c<0$ then $1-c>0$ so $c(1-c)<0$
answered Sep 1 at 7:34
greedoid
28k93776
28k93776
add a comment |Â
add a comment |Â
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f2901440%2fprove-that-abc-le-0%23new-answer', 'question_page');
);
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password