Proof of an inequality: is it correct?
$begingroup$
Let $x_{1},cdots, x_{n}>-1$ be real numbers such that
$sum{x_{i}}=n$. Prove that: $$sum_{i=1}^{n}{frac{1}{x_{i}+1}}geq sum_{i=1}^{n}{frac{x_{i}}{x_{i}^{2}+1}}$$
My proof: By AM-HM and $sum{x_{i}}=n$, LHS $geq frac{n^{2}}{n+ntimes 1}=frac{n}{2}$
On the other hand, by using AM-GM in the denominator, RHS $leq sum{frac{x_{i}}{2x_{i}}}=frac{n}{2} rightarrow$ LHS $geq$ RHS, with equality iff for all $i$, $x_{i}=1$.
Is this proof correct?
proof-verification inequality a.m.-g.m.-inequality cauchy-schwarz-inequality tangent-line-method
edited Dec 6 '18 at 14:42
Michael Rozenberg
98.3k1590188
$endgroup$
add a comment |
$begingroup$
Let $x_{1},cdots, x_{n}>-1$ be real numbers such that
$sum{x_{i}}=n$. Prove that: $$sum_{i=1}^{n}{frac{1}{x_{i}+1}}geq sum_{i=1}^{n}{frac{x_{i}}{x_{i}^{2}+1}}$$
My proof: By AM-HM and $sum{x_{i}}=n$, LHS $geq frac{n^{2}}{n+ntimes 1}=frac{n}{2}$
On the other hand, by using AM-GM in the denominator, RHS $leq sum{frac{x_{i}}{2x_{i}}}=frac{n}{2} rightarrow$ LHS $geq$ RHS, with equality iff for all $i$, $x_{i}=1$.
Is this proof correct?
proof-verification inequality a.m.-g.m.-inequality cauchy-schwarz-inequality tangent-line-method
edited Dec 6 '18 at 14:42
Michael Rozenberg
98.3k1590188
$endgroup$
$begingroup$
Yes, it’s correct.
$endgroup$
– Macavity
Dec 11 '18 at 12:25
add a comment |
$begingroup$
Let $x_{1},cdots, x_{n}>-1$ be real numbers such that
$sum{x_{i}}=n$. Prove that: $$sum_{i=1}^{n}{frac{1}{x_{i}+1}}geq sum_{i=1}^{n}{frac{x_{i}}{x_{i}^{2}+1}}$$
My proof: By AM-HM and $sum{x_{i}}=n$, LHS $geq frac{n^{2}}{n+ntimes 1}=frac{n}{2}$
On the other hand, by using AM-GM in the denominator, RHS $leq sum{frac{x_{i}}{2x_{i}}}=frac{n}{2} rightarrow$ LHS $geq$ RHS, with equality iff for all $i$, $x_{i}=1$.
Is this proof correct?
proof-verification inequality a.m.-g.m.-inequality cauchy-schwarz-inequality tangent-line-method
edited Dec 6 '18 at 14:42
Michael Rozenberg
98.3k1590188
$endgroup$
Let $x_{1},cdots, x_{n}>-1$ be real numbers such that
$sum{x_{i}}=n$. Prove that: $$sum_{i=1}^{n}{frac{1}{x_{i}+1}}geq sum_{i=1}^{n}{frac{x_{i}}{x_{i}^{2}+1}}$$
My proof: By AM-HM and $sum{x_{i}}=n$, LHS $geq frac{n^{2}}{n+ntimes 1}=frac{n}{2}$
On the other hand, by using AM-GM in the denominator, RHS $leq sum{frac{x_{i}}{2x_{i}}}=frac{n}{2} rightarrow$ LHS $geq$ RHS, with equality iff for all $i$, $x_{i}=1$.
Is this proof correct?
proof-verification inequality a.m.-g.m.-inequality cauchy-schwarz-inequality tangent-line-method
proof-verification inequality a.m.-g.m.-inequality cauchy-schwarz-inequality tangent-line-method
edited Dec 6 '18 at 14:42
Michael Rozenberg
98.3k1590188
edited Dec 6 '18 at 14:42
Michael Rozenberg
98.3k1590188
edited Dec 6 '18 at 14:42
Michael Rozenberg
98.3k1590188
edited Dec 6 '18 at 14:42
Michael Rozenberg
98.3k1590188
edited Dec 6 '18 at 14:42
Michael Rozenberg
98.3k1590188
98.3k1590188
asked Dec 6 '18 at 14:06
user542970
$begingroup$
Yes, it’s correct.
$endgroup$
– Macavity
Dec 11 '18 at 12:25
add a comment |
$begingroup$
Yes, it’s correct.
$endgroup$
– Macavity
Dec 11 '18 at 12:25
$begingroup$
Yes, it’s correct.
$endgroup$
– Macavity
Dec 11 '18 at 12:25
$begingroup$
Yes, it’s correct.
$endgroup$
– Macavity
Dec 11 '18 at 12:25
add a comment |
1 Answer
1
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oldest
votes
$begingroup$
Your first step is right.
Now, your inequality follows from the following C-S.
$$sum_{i=1}^n(1+x_i)sum_{cyc}frac{1}{1+x_i}geq n^2.$$
Also, we can use the TL method:
$$sum_{i=1}^nleft(frac{1}{1+x_i}-frac{1}{1+x_i^2}right)=sum_{i=1}^nfrac{1-x_i}{(1+x_i)(1+x_i^2)}=$$
$$=sum_{i=1}^nleft(frac{1-x_i}{(1+x_i)(1+x_i^2)}+frac{1}{4}(x_i-1)right)=sum_{i=1}^nfrac{(x_i-1)^2(x_i^2+2x_i+3)}{4(1+x_i)(1+x_i^2)}geq0.$$
answered Dec 6 '18 at 14:41
Michael RozenbergMichael Rozenberg
98.3k1590188
$endgroup$
$begingroup$
I was at that first step but I can't go further! Sorry for the (maybe) stupid question, but what is TL method?
$endgroup$
– gimusi
Dec 6 '18 at 14:44
$begingroup$
Because $sum (x_i-1)=0$?
$endgroup$
– gimusi
Dec 6 '18 at 14:45
$begingroup$
@gimusi Yes, of course! TL it's the Tangent Line method. See here: math.stackexchange.com/tags/tangent-line-method/info
$endgroup$
– Michael Rozenberg
Dec 6 '18 at 14:45
$begingroup$
What was wrong with the second step?
$endgroup$
– user542970
Dec 6 '18 at 14:48
1
$begingroup$
Yes, of course. Now, I see this point. Id est, your solution is right!
$endgroup$
– Michael Rozenberg
Dec 6 '18 at 14:57
|
show 6 more comments
Your Answer
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1 Answer
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1 Answer
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active
oldest
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active
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$begingroup$
Your first step is right.
Now, your inequality follows from the following C-S.
$$sum_{i=1}^n(1+x_i)sum_{cyc}frac{1}{1+x_i}geq n^2.$$
Also, we can use the TL method:
$$sum_{i=1}^nleft(frac{1}{1+x_i}-frac{1}{1+x_i^2}right)=sum_{i=1}^nfrac{1-x_i}{(1+x_i)(1+x_i^2)}=$$
$$=sum_{i=1}^nleft(frac{1-x_i}{(1+x_i)(1+x_i^2)}+frac{1}{4}(x_i-1)right)=sum_{i=1}^nfrac{(x_i-1)^2(x_i^2+2x_i+3)}{4(1+x_i)(1+x_i^2)}geq0.$$
answered Dec 6 '18 at 14:41
Michael RozenbergMichael Rozenberg
98.3k1590188
$endgroup$
$begingroup$
I was at that first step but I can't go further! Sorry for the (maybe) stupid question, but what is TL method?
$endgroup$
– gimusi
Dec 6 '18 at 14:44
$begingroup$
Because $sum (x_i-1)=0$?
$endgroup$
– gimusi
Dec 6 '18 at 14:45
$begingroup$
@gimusi Yes, of course! TL it's the Tangent Line method. See here: math.stackexchange.com/tags/tangent-line-method/info
$endgroup$
– Michael Rozenberg
Dec 6 '18 at 14:45
$begingroup$
What was wrong with the second step?
$endgroup$
– user542970
Dec 6 '18 at 14:48
1
$begingroup$
Yes, of course. Now, I see this point. Id est, your solution is right!
$endgroup$
– Michael Rozenberg
Dec 6 '18 at 14:57
|
show 6 more comments
$begingroup$
Your first step is right.
Now, your inequality follows from the following C-S.
$$sum_{i=1}^n(1+x_i)sum_{cyc}frac{1}{1+x_i}geq n^2.$$
Also, we can use the TL method:
$$sum_{i=1}^nleft(frac{1}{1+x_i}-frac{1}{1+x_i^2}right)=sum_{i=1}^nfrac{1-x_i}{(1+x_i)(1+x_i^2)}=$$
$$=sum_{i=1}^nleft(frac{1-x_i}{(1+x_i)(1+x_i^2)}+frac{1}{4}(x_i-1)right)=sum_{i=1}^nfrac{(x_i-1)^2(x_i^2+2x_i+3)}{4(1+x_i)(1+x_i^2)}geq0.$$
answered Dec 6 '18 at 14:41
Michael RozenbergMichael Rozenberg
98.3k1590188
$endgroup$
$begingroup$
I was at that first step but I can't go further! Sorry for the (maybe) stupid question, but what is TL method?
$endgroup$
– gimusi
Dec 6 '18 at 14:44
$begingroup$
Because $sum (x_i-1)=0$?
$endgroup$
– gimusi
Dec 6 '18 at 14:45
$begingroup$
@gimusi Yes, of course! TL it's the Tangent Line method. See here: math.stackexchange.com/tags/tangent-line-method/info
$endgroup$
– Michael Rozenberg
Dec 6 '18 at 14:45
$begingroup$
What was wrong with the second step?
$endgroup$
– user542970
Dec 6 '18 at 14:48
1
$begingroup$
Yes, of course. Now, I see this point. Id est, your solution is right!
$endgroup$
– Michael Rozenberg
Dec 6 '18 at 14:57
|
show 6 more comments
$begingroup$
Your first step is right.
Now, your inequality follows from the following C-S.
$$sum_{i=1}^n(1+x_i)sum_{cyc}frac{1}{1+x_i}geq n^2.$$
Also, we can use the TL method:
$$sum_{i=1}^nleft(frac{1}{1+x_i}-frac{1}{1+x_i^2}right)=sum_{i=1}^nfrac{1-x_i}{(1+x_i)(1+x_i^2)}=$$
$$=sum_{i=1}^nleft(frac{1-x_i}{(1+x_i)(1+x_i^2)}+frac{1}{4}(x_i-1)right)=sum_{i=1}^nfrac{(x_i-1)^2(x_i^2+2x_i+3)}{4(1+x_i)(1+x_i^2)}geq0.$$
answered Dec 6 '18 at 14:41
Michael RozenbergMichael Rozenberg
98.3k1590188
$endgroup$
Your first step is right.
Now, your inequality follows from the following C-S.
$$sum_{i=1}^n(1+x_i)sum_{cyc}frac{1}{1+x_i}geq n^2.$$
Also, we can use the TL method:
$$sum_{i=1}^nleft(frac{1}{1+x_i}-frac{1}{1+x_i^2}right)=sum_{i=1}^nfrac{1-x_i}{(1+x_i)(1+x_i^2)}=$$
$$=sum_{i=1}^nleft(frac{1-x_i}{(1+x_i)(1+x_i^2)}+frac{1}{4}(x_i-1)right)=sum_{i=1}^nfrac{(x_i-1)^2(x_i^2+2x_i+3)}{4(1+x_i)(1+x_i^2)}geq0.$$
answered Dec 6 '18 at 14:41
Michael RozenbergMichael Rozenberg
98.3k1590188
answered Dec 6 '18 at 14:41
Michael RozenbergMichael Rozenberg
98.3k1590188
answered Dec 6 '18 at 14:41
Michael RozenbergMichael Rozenberg
98.3k1590188
answered Dec 6 '18 at 14:41
Michael RozenbergMichael Rozenberg
98.3k1590188
98.3k1590188
$begingroup$
I was at that first step but I can't go further! Sorry for the (maybe) stupid question, but what is TL method?
$endgroup$
– gimusi
Dec 6 '18 at 14:44
$begingroup$
Because $sum (x_i-1)=0$?
$endgroup$
– gimusi
Dec 6 '18 at 14:45
$begingroup$
@gimusi Yes, of course! TL it's the Tangent Line method. See here: math.stackexchange.com/tags/tangent-line-method/info
$endgroup$
– Michael Rozenberg
Dec 6 '18 at 14:45
$begingroup$
What was wrong with the second step?
$endgroup$
– user542970
Dec 6 '18 at 14:48
1
$begingroup$
Yes, of course. Now, I see this point. Id est, your solution is right!
$endgroup$
– Michael Rozenberg
Dec 6 '18 at 14:57
|
show 6 more comments
$begingroup$
I was at that first step but I can't go further! Sorry for the (maybe) stupid question, but what is TL method?
$endgroup$
– gimusi
Dec 6 '18 at 14:44
$begingroup$
Because $sum (x_i-1)=0$?
$endgroup$
– gimusi
Dec 6 '18 at 14:45
$begingroup$
@gimusi Yes, of course! TL it's the Tangent Line method. See here: math.stackexchange.com/tags/tangent-line-method/info
$endgroup$
– Michael Rozenberg
Dec 6 '18 at 14:45
$begingroup$
What was wrong with the second step?
$endgroup$
– user542970
Dec 6 '18 at 14:48
1
$begingroup$
Yes, of course. Now, I see this point. Id est, your solution is right!
$endgroup$
– Michael Rozenberg
Dec 6 '18 at 14:57
$begingroup$
I was at that first step but I can't go further! Sorry for the (maybe) stupid question, but what is TL method?
$endgroup$
– gimusi
Dec 6 '18 at 14:44
$begingroup$
I was at that first step but I can't go further! Sorry for the (maybe) stupid question, but what is TL method?
$endgroup$
– gimusi
Dec 6 '18 at 14:44
$begingroup$
Because $sum (x_i-1)=0$?
$endgroup$
– gimusi
Dec 6 '18 at 14:45
$begingroup$
Because $sum (x_i-1)=0$?
$endgroup$
– gimusi
Dec 6 '18 at 14:45
$begingroup$
@gimusi Yes, of course! TL it's the Tangent Line method. See here: math.stackexchange.com/tags/tangent-line-method/info
$endgroup$
– Michael Rozenberg
Dec 6 '18 at 14:45
$begingroup$
@gimusi Yes, of course! TL it's the Tangent Line method. See here: math.stackexchange.com/tags/tangent-line-method/info
$endgroup$
– Michael Rozenberg
Dec 6 '18 at 14:45
$begingroup$
What was wrong with the second step?
$endgroup$
– user542970
Dec 6 '18 at 14:48
$begingroup$
What was wrong with the second step?
$endgroup$
– user542970
Dec 6 '18 at 14:48
1
1
$begingroup$
Yes, of course. Now, I see this point. Id est, your solution is right!
$endgroup$
– Michael Rozenberg
Dec 6 '18 at 14:57
$begingroup$
Yes, of course. Now, I see this point. Id est, your solution is right!
$endgroup$
– Michael Rozenberg
Dec 6 '18 at 14:57
|
show 6 more comments
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$begingroup$
Yes, it’s correct.
$endgroup$
– Macavity
Dec 11 '18 at 12:25