Proof of an inequality: is it correct?












2












$begingroup$



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







share|cite|improve this question











$endgroup$












  • $begingroup$
    Yes, it’s correct.
    $endgroup$
    – Macavity
    Dec 11 '18 at 12:25
















2












$begingroup$



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







share|cite|improve this question











$endgroup$












  • $begingroup$
    Yes, it’s correct.
    $endgroup$
    – Macavity
    Dec 11 '18 at 12:25














2












2








2


1



$begingroup$



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







share|cite|improve this question











$endgroup$





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






share|cite|improve this question















share|cite|improve this question













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share|cite|improve this question








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


















  • $begingroup$
    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










1 Answer
1






active

oldest

votes


















1












$begingroup$

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:
$$sum_{i=1}^nleft(frac{1}{1+x_i}-frac{1}{1+x_i^2}right)=sum_{i=1}^nfrac{1-x_i}{(1+x_i)(1+x_i^2)}=$$
$$=sum_{i=1}^nleft(frac{1-x_i}{(1+x_i)(1+x_i^2)}+frac{1}{4}(x_i-1)right)=sum_{i=1}^nfrac{(x_i-1)^2(x_i^2+2x_i+3)}{4(1+x_i)(1+x_i^2)}geq0.$$






share|cite|improve this answer









$endgroup$













  • $begingroup$
    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













Your Answer





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1 Answer
1






active

oldest

votes








1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes









1












$begingroup$

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:
$$sum_{i=1}^nleft(frac{1}{1+x_i}-frac{1}{1+x_i^2}right)=sum_{i=1}^nfrac{1-x_i}{(1+x_i)(1+x_i^2)}=$$
$$=sum_{i=1}^nleft(frac{1-x_i}{(1+x_i)(1+x_i^2)}+frac{1}{4}(x_i-1)right)=sum_{i=1}^nfrac{(x_i-1)^2(x_i^2+2x_i+3)}{4(1+x_i)(1+x_i^2)}geq0.$$






share|cite|improve this answer









$endgroup$













  • $begingroup$
    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


















1












$begingroup$

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:
$$sum_{i=1}^nleft(frac{1}{1+x_i}-frac{1}{1+x_i^2}right)=sum_{i=1}^nfrac{1-x_i}{(1+x_i)(1+x_i^2)}=$$
$$=sum_{i=1}^nleft(frac{1-x_i}{(1+x_i)(1+x_i^2)}+frac{1}{4}(x_i-1)right)=sum_{i=1}^nfrac{(x_i-1)^2(x_i^2+2x_i+3)}{4(1+x_i)(1+x_i^2)}geq0.$$






share|cite|improve this answer









$endgroup$













  • $begingroup$
    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
















1












1








1





$begingroup$

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:
$$sum_{i=1}^nleft(frac{1}{1+x_i}-frac{1}{1+x_i^2}right)=sum_{i=1}^nfrac{1-x_i}{(1+x_i)(1+x_i^2)}=$$
$$=sum_{i=1}^nleft(frac{1-x_i}{(1+x_i)(1+x_i^2)}+frac{1}{4}(x_i-1)right)=sum_{i=1}^nfrac{(x_i-1)^2(x_i^2+2x_i+3)}{4(1+x_i)(1+x_i^2)}geq0.$$






share|cite|improve this answer









$endgroup$



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:
$$sum_{i=1}^nleft(frac{1}{1+x_i}-frac{1}{1+x_i^2}right)=sum_{i=1}^nfrac{1-x_i}{(1+x_i)(1+x_i^2)}=$$
$$=sum_{i=1}^nleft(frac{1-x_i}{(1+x_i)(1+x_i^2)}+frac{1}{4}(x_i-1)right)=sum_{i=1}^nfrac{(x_i-1)^2(x_i^2+2x_i+3)}{4(1+x_i)(1+x_i^2)}geq0.$$







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










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




















  • $begingroup$
    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




















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