Prove that $sqrt{1+a} + sqrt{1+b} + sqrt{1+c} leq 4$ [closed]
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I am trying to prove this inequality but I am having some difficulties.
$sqrt{1+a} + sqrt{1+b} + sqrt{1+c} leq 4$
Edit: sorry i forgot to add a crucial info:
$a + b + c = 2 $ and $a,b,c in R+$
arithmetic proof-theory
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closed as off-topic by Carl Mummert, Shailesh, user91500, Lee David Chung Lin, José Carlos Santos Jan 20 at 11:43
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Carl Mummert, Shailesh, user91500, Lee David Chung Lin, José Carlos Santos
If this question can be reworded to fit the rules in the help center, please edit the question.
add a comment |
$begingroup$
I am trying to prove this inequality but I am having some difficulties.
$sqrt{1+a} + sqrt{1+b} + sqrt{1+c} leq 4$
Edit: sorry i forgot to add a crucial info:
$a + b + c = 2 $ and $a,b,c in R+$
arithmetic proof-theory
$endgroup$
closed as off-topic by Carl Mummert, Shailesh, user91500, Lee David Chung Lin, José Carlos Santos Jan 20 at 11:43
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Carl Mummert, Shailesh, user91500, Lee David Chung Lin, José Carlos Santos
If this question can be reworded to fit the rules in the help center, please edit the question.
4
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The result is obviously false without some restriction on $a$, $b$, and $c$.
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– user3482749
Jan 19 at 22:36
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@user3482749 sorry about that , i forgot to add some info
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– souleatzz
Jan 19 at 22:42
add a comment |
$begingroup$
I am trying to prove this inequality but I am having some difficulties.
$sqrt{1+a} + sqrt{1+b} + sqrt{1+c} leq 4$
Edit: sorry i forgot to add a crucial info:
$a + b + c = 2 $ and $a,b,c in R+$
arithmetic proof-theory
$endgroup$
I am trying to prove this inequality but I am having some difficulties.
$sqrt{1+a} + sqrt{1+b} + sqrt{1+c} leq 4$
Edit: sorry i forgot to add a crucial info:
$a + b + c = 2 $ and $a,b,c in R+$
arithmetic proof-theory
arithmetic proof-theory
edited Jan 19 at 22:41
souleatzz
asked Jan 19 at 22:34
souleatzzsouleatzz
32
32
closed as off-topic by Carl Mummert, Shailesh, user91500, Lee David Chung Lin, José Carlos Santos Jan 20 at 11:43
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Carl Mummert, Shailesh, user91500, Lee David Chung Lin, José Carlos Santos
If this question can be reworded to fit the rules in the help center, please edit the question.
closed as off-topic by Carl Mummert, Shailesh, user91500, Lee David Chung Lin, José Carlos Santos Jan 20 at 11:43
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Carl Mummert, Shailesh, user91500, Lee David Chung Lin, José Carlos Santos
If this question can be reworded to fit the rules in the help center, please edit the question.
4
$begingroup$
The result is obviously false without some restriction on $a$, $b$, and $c$.
$endgroup$
– user3482749
Jan 19 at 22:36
$begingroup$
@user3482749 sorry about that , i forgot to add some info
$endgroup$
– souleatzz
Jan 19 at 22:42
add a comment |
4
$begingroup$
The result is obviously false without some restriction on $a$, $b$, and $c$.
$endgroup$
– user3482749
Jan 19 at 22:36
$begingroup$
@user3482749 sorry about that , i forgot to add some info
$endgroup$
– souleatzz
Jan 19 at 22:42
4
4
$begingroup$
The result is obviously false without some restriction on $a$, $b$, and $c$.
$endgroup$
– user3482749
Jan 19 at 22:36
$begingroup$
The result is obviously false without some restriction on $a$, $b$, and $c$.
$endgroup$
– user3482749
Jan 19 at 22:36
$begingroup$
@user3482749 sorry about that , i forgot to add some info
$endgroup$
– souleatzz
Jan 19 at 22:42
$begingroup$
@user3482749 sorry about that , i forgot to add some info
$endgroup$
– souleatzz
Jan 19 at 22:42
add a comment |
1 Answer
1
active
oldest
votes
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Hint: Use Bernoulli's inequality: for $xgeqslant-1$ and $0 leqslant r leqslant 1$, we have that
$$(1+x)^r leqslant 1+rx$$
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Great, thank you. Pretty easy after your help.
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– souleatzz
Jan 19 at 22:53
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@souleatzz I'm glad I could help :) But next time please try to include your work in the question, even if you think that it's wrong :)
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– Botond
Jan 19 at 22:56
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yeah, sorry about that. I actually did try it but it was totally wrong and pretty long. Thank you again.
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– souleatzz
Jan 19 at 22:58
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Hint: Use Bernoulli's inequality: for $xgeqslant-1$ and $0 leqslant r leqslant 1$, we have that
$$(1+x)^r leqslant 1+rx$$
$endgroup$
$begingroup$
Great, thank you. Pretty easy after your help.
$endgroup$
– souleatzz
Jan 19 at 22:53
$begingroup$
@souleatzz I'm glad I could help :) But next time please try to include your work in the question, even if you think that it's wrong :)
$endgroup$
– Botond
Jan 19 at 22:56
$begingroup$
yeah, sorry about that. I actually did try it but it was totally wrong and pretty long. Thank you again.
$endgroup$
– souleatzz
Jan 19 at 22:58
add a comment |
$begingroup$
Hint: Use Bernoulli's inequality: for $xgeqslant-1$ and $0 leqslant r leqslant 1$, we have that
$$(1+x)^r leqslant 1+rx$$
$endgroup$
$begingroup$
Great, thank you. Pretty easy after your help.
$endgroup$
– souleatzz
Jan 19 at 22:53
$begingroup$
@souleatzz I'm glad I could help :) But next time please try to include your work in the question, even if you think that it's wrong :)
$endgroup$
– Botond
Jan 19 at 22:56
$begingroup$
yeah, sorry about that. I actually did try it but it was totally wrong and pretty long. Thank you again.
$endgroup$
– souleatzz
Jan 19 at 22:58
add a comment |
$begingroup$
Hint: Use Bernoulli's inequality: for $xgeqslant-1$ and $0 leqslant r leqslant 1$, we have that
$$(1+x)^r leqslant 1+rx$$
$endgroup$
Hint: Use Bernoulli's inequality: for $xgeqslant-1$ and $0 leqslant r leqslant 1$, we have that
$$(1+x)^r leqslant 1+rx$$
edited Jan 19 at 22:56
answered Jan 19 at 22:50
BotondBotond
5,9252832
5,9252832
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Great, thank you. Pretty easy after your help.
$endgroup$
– souleatzz
Jan 19 at 22:53
$begingroup$
@souleatzz I'm glad I could help :) But next time please try to include your work in the question, even if you think that it's wrong :)
$endgroup$
– Botond
Jan 19 at 22:56
$begingroup$
yeah, sorry about that. I actually did try it but it was totally wrong and pretty long. Thank you again.
$endgroup$
– souleatzz
Jan 19 at 22:58
add a comment |
$begingroup$
Great, thank you. Pretty easy after your help.
$endgroup$
– souleatzz
Jan 19 at 22:53
$begingroup$
@souleatzz I'm glad I could help :) But next time please try to include your work in the question, even if you think that it's wrong :)
$endgroup$
– Botond
Jan 19 at 22:56
$begingroup$
yeah, sorry about that. I actually did try it but it was totally wrong and pretty long. Thank you again.
$endgroup$
– souleatzz
Jan 19 at 22:58
$begingroup$
Great, thank you. Pretty easy after your help.
$endgroup$
– souleatzz
Jan 19 at 22:53
$begingroup$
Great, thank you. Pretty easy after your help.
$endgroup$
– souleatzz
Jan 19 at 22:53
$begingroup$
@souleatzz I'm glad I could help :) But next time please try to include your work in the question, even if you think that it's wrong :)
$endgroup$
– Botond
Jan 19 at 22:56
$begingroup$
@souleatzz I'm glad I could help :) But next time please try to include your work in the question, even if you think that it's wrong :)
$endgroup$
– Botond
Jan 19 at 22:56
$begingroup$
yeah, sorry about that. I actually did try it but it was totally wrong and pretty long. Thank you again.
$endgroup$
– souleatzz
Jan 19 at 22:58
$begingroup$
yeah, sorry about that. I actually did try it but it was totally wrong and pretty long. Thank you again.
$endgroup$
– souleatzz
Jan 19 at 22:58
add a comment |
4
$begingroup$
The result is obviously false without some restriction on $a$, $b$, and $c$.
$endgroup$
– user3482749
Jan 19 at 22:36
$begingroup$
@user3482749 sorry about that , i forgot to add some info
$endgroup$
– souleatzz
Jan 19 at 22:42