An inequality for convex function from $mathbb R$ to $mathbb R$












1












$begingroup$


If $f(x): mathbb R to mathbb R$ is a convex function, prove that
$$f(x_1) + f(x_2) + f(x_3) + 3 f(frac{x_1 + x_2 + x_3}{3}) geq 2 f(frac{x_1 + x_2}{2}) + 2 f(frac{x_1 + x_3}{2}) + 2 f(frac{x_2 + x_3}{2})$$



What I have tried:
If two of $x_1$, $x_2$, $x_3$ are equal, without loss of generality, assuming $x_2 = x_3$, then the inequality degenerate to
begin{align}
f(x_1) + 3 f(frac{x_1 + 2 x_2}{3}) geq 4 f(frac{x_1 + x_2}{2})
end{align}
Since $f$ is convex, we have
$
frac{1}{4} f(x_1) + frac{3}{4} f(frac{x_1 + 2 x_2}{3})
geq f(frac{1}{4} x_1 + frac{3}{4} frac{x_1 + 2 x_2}{3})
geq f(frac{x_1 + x_2}{2})
$, thus the inequality holds.



Then we consider the case where $x_1$, $x_2$, $x_3$ are distinct. Without loss of generality, we assume $0 = x_1 < x_2 = lambda leq frac{1}{2} < x_3 = 1$. Then the inequality degenerates to
begin{align}
frac{1}{2} f(0) + frac{1}{2} f(lambda) + frac{3}{2} f(frac{lambda + 1}{3}) + frac{1}{2} f(1) geq f(frac{lambda}{2}) + f(frac{1}{2}) + f(frac{lambda + 1}{2})
end{align}
If $x_2 = frac{1}{2}$, it further degenerates to
$frac{1}{2} f(0) + frac{1}{2} f(0.5) + frac{1}{2} f(0.5) + frac{1}{2} f(1) geq f(0.25) + f(0.75),$
which is obviously true. Thus we can only consider $0 < lambda < frac{1}{2}$ (then $0 < frac{lambda}{2} < lambda < frac{lambda + 1}{3} < frac{1}{2} < frac{lambda + 1}{2} < 1$).






convex-analysis convex-optimization jensen-inequality







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$endgroup$








  • 1




    $begingroup$
    This is known as Popoviciu's inequality
    $endgroup$
    – munchhausen
    Dec 19 '18 at 14:49
















1












$begingroup$


If $f(x): mathbb R to mathbb R$ is a convex function, prove that
$$f(x_1) + f(x_2) + f(x_3) + 3 f(frac{x_1 + x_2 + x_3}{3}) geq 2 f(frac{x_1 + x_2}{2}) + 2 f(frac{x_1 + x_3}{2}) + 2 f(frac{x_2 + x_3}{2})$$



What I have tried:
If two of $x_1$, $x_2$, $x_3$ are equal, without loss of generality, assuming $x_2 = x_3$, then the inequality degenerate to
begin{align}
f(x_1) + 3 f(frac{x_1 + 2 x_2}{3}) geq 4 f(frac{x_1 + x_2}{2})
end{align}
Since $f$ is convex, we have
$
frac{1}{4} f(x_1) + frac{3}{4} f(frac{x_1 + 2 x_2}{3})
geq f(frac{1}{4} x_1 + frac{3}{4} frac{x_1 + 2 x_2}{3})
geq f(frac{x_1 + x_2}{2})
$, thus the inequality holds.



Then we consider the case where $x_1$, $x_2$, $x_3$ are distinct. Without loss of generality, we assume $0 = x_1 < x_2 = lambda leq frac{1}{2} < x_3 = 1$. Then the inequality degenerates to
begin{align}
frac{1}{2} f(0) + frac{1}{2} f(lambda) + frac{3}{2} f(frac{lambda + 1}{3}) + frac{1}{2} f(1) geq f(frac{lambda}{2}) + f(frac{1}{2}) + f(frac{lambda + 1}{2})
end{align}
If $x_2 = frac{1}{2}$, it further degenerates to
$frac{1}{2} f(0) + frac{1}{2} f(0.5) + frac{1}{2} f(0.5) + frac{1}{2} f(1) geq f(0.25) + f(0.75),$
which is obviously true. Thus we can only consider $0 < lambda < frac{1}{2}$ (then $0 < frac{lambda}{2} < lambda < frac{lambda + 1}{3} < frac{1}{2} < frac{lambda + 1}{2} < 1$).






convex-analysis convex-optimization jensen-inequality







share|cite|improve this question









$endgroup$








  • 1




    $begingroup$
    This is known as Popoviciu's inequality
    $endgroup$
    – munchhausen
    Dec 19 '18 at 14:49














1












1








1


1



$begingroup$


If $f(x): mathbb R to mathbb R$ is a convex function, prove that
$$f(x_1) + f(x_2) + f(x_3) + 3 f(frac{x_1 + x_2 + x_3}{3}) geq 2 f(frac{x_1 + x_2}{2}) + 2 f(frac{x_1 + x_3}{2}) + 2 f(frac{x_2 + x_3}{2})$$



What I have tried:
If two of $x_1$, $x_2$, $x_3$ are equal, without loss of generality, assuming $x_2 = x_3$, then the inequality degenerate to
begin{align}
f(x_1) + 3 f(frac{x_1 + 2 x_2}{3}) geq 4 f(frac{x_1 + x_2}{2})
end{align}
Since $f$ is convex, we have
$
frac{1}{4} f(x_1) + frac{3}{4} f(frac{x_1 + 2 x_2}{3})
geq f(frac{1}{4} x_1 + frac{3}{4} frac{x_1 + 2 x_2}{3})
geq f(frac{x_1 + x_2}{2})
$, thus the inequality holds.



Then we consider the case where $x_1$, $x_2$, $x_3$ are distinct. Without loss of generality, we assume $0 = x_1 < x_2 = lambda leq frac{1}{2} < x_3 = 1$. Then the inequality degenerates to
begin{align}
frac{1}{2} f(0) + frac{1}{2} f(lambda) + frac{3}{2} f(frac{lambda + 1}{3}) + frac{1}{2} f(1) geq f(frac{lambda}{2}) + f(frac{1}{2}) + f(frac{lambda + 1}{2})
end{align}
If $x_2 = frac{1}{2}$, it further degenerates to
$frac{1}{2} f(0) + frac{1}{2} f(0.5) + frac{1}{2} f(0.5) + frac{1}{2} f(1) geq f(0.25) + f(0.75),$
which is obviously true. Thus we can only consider $0 < lambda < frac{1}{2}$ (then $0 < frac{lambda}{2} < lambda < frac{lambda + 1}{3} < frac{1}{2} < frac{lambda + 1}{2} < 1$).






convex-analysis convex-optimization jensen-inequality







share|cite|improve this question









$endgroup$




If $f(x): mathbb R to mathbb R$ is a convex function, prove that
$$f(x_1) + f(x_2) + f(x_3) + 3 f(frac{x_1 + x_2 + x_3}{3}) geq 2 f(frac{x_1 + x_2}{2}) + 2 f(frac{x_1 + x_3}{2}) + 2 f(frac{x_2 + x_3}{2})$$



What I have tried:
If two of $x_1$, $x_2$, $x_3$ are equal, without loss of generality, assuming $x_2 = x_3$, then the inequality degenerate to
begin{align}
f(x_1) + 3 f(frac{x_1 + 2 x_2}{3}) geq 4 f(frac{x_1 + x_2}{2})
end{align}
Since $f$ is convex, we have
$
frac{1}{4} f(x_1) + frac{3}{4} f(frac{x_1 + 2 x_2}{3})
geq f(frac{1}{4} x_1 + frac{3}{4} frac{x_1 + 2 x_2}{3})
geq f(frac{x_1 + x_2}{2})
$, thus the inequality holds.



Then we consider the case where $x_1$, $x_2$, $x_3$ are distinct. Without loss of generality, we assume $0 = x_1 < x_2 = lambda leq frac{1}{2} < x_3 = 1$. Then the inequality degenerates to
begin{align}
frac{1}{2} f(0) + frac{1}{2} f(lambda) + frac{3}{2} f(frac{lambda + 1}{3}) + frac{1}{2} f(1) geq f(frac{lambda}{2}) + f(frac{1}{2}) + f(frac{lambda + 1}{2})
end{align}
If $x_2 = frac{1}{2}$, it further degenerates to
$frac{1}{2} f(0) + frac{1}{2} f(0.5) + frac{1}{2} f(0.5) + frac{1}{2} f(1) geq f(0.25) + f(0.75),$
which is obviously true. Thus we can only consider $0 < lambda < frac{1}{2}$ (then $0 < frac{lambda}{2} < lambda < frac{lambda + 1}{3} < frac{1}{2} < frac{lambda + 1}{2} < 1$).






convex-analysis convex-optimization jensen-inequality




convex-analysis convex-optimization jensen-inequality






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asked Dec 19 '18 at 14:13









Rubisco LeeRubisco Lee

1148




1148








  • 1




    $begingroup$
    This is known as Popoviciu's inequality
    $endgroup$
    – munchhausen
    Dec 19 '18 at 14:49














  • 1




    $begingroup$
    This is known as Popoviciu's inequality
    $endgroup$
    – munchhausen
    Dec 19 '18 at 14:49








1




1




$begingroup$
This is known as Popoviciu's inequality
$endgroup$
– munchhausen
Dec 19 '18 at 14:49




$begingroup$
This is known as Popoviciu's inequality
$endgroup$
– munchhausen
Dec 19 '18 at 14:49










1 Answer
1






active

oldest

votes


















3












$begingroup$

A proof of Popoviciu's inequality follows from Jensen's Inequality for convex functions. Wlog, assume $x le y le z$ there are two possibilities. Either $displaystyle y < frac{x+y+z}{3}$ or the reverse.



In the first case, $displaystyle frac{x+y+z}{3} < frac{x+z}{2}< z$ and $displaystyle frac{x+y+z}{3} < frac{y+z}{2} < z$, thus, there are positive real numbers $lambda_1,lambda_2 >0$ such that,



$displaystyle frac{x+z}{2} = lambda_1 z + (1-lambda_1)left(frac{x+y+z}{3}right)$ and $displaystyle frac{y+z}{2} = lambda_2 z + (1-lambda_2)left(frac{x+y+z}{3}right)$



It is easy to verify that $lambda_1 + lambda_2 = dfrac{1}{2}$.



Now, using Jensen's inequality one has:



$displaystyle fleft(frac{x+y}{2}right) le frac{f(x)+f(y)}{2}$



$displaystyle fleft(frac{x+z}{2}right) = lambda_1 f(z) + (1-lambda_1)fleft(frac{x+y+z}{3}right)$



$displaystyle fleft(frac{y+z}{2}right) = lambda_2 f(z) + (1-lambda_2)fleft(frac{x+y+z}{3}right)$



Adding the three inequalities gives the Popoviciu's inequality. The second case where $displaystyle y > frac{x+y+z}{3}$ we simply interchange the roles of $x$ and $z$ and the rest is similar.






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    $begingroup$

    A proof of Popoviciu's inequality follows from Jensen's Inequality for convex functions. Wlog, assume $x le y le z$ there are two possibilities. Either $displaystyle y < frac{x+y+z}{3}$ or the reverse.



    In the first case, $displaystyle frac{x+y+z}{3} < frac{x+z}{2}< z$ and $displaystyle frac{x+y+z}{3} < frac{y+z}{2} < z$, thus, there are positive real numbers $lambda_1,lambda_2 >0$ such that,



    $displaystyle frac{x+z}{2} = lambda_1 z + (1-lambda_1)left(frac{x+y+z}{3}right)$ and $displaystyle frac{y+z}{2} = lambda_2 z + (1-lambda_2)left(frac{x+y+z}{3}right)$



    It is easy to verify that $lambda_1 + lambda_2 = dfrac{1}{2}$.



    Now, using Jensen's inequality one has:



    $displaystyle fleft(frac{x+y}{2}right) le frac{f(x)+f(y)}{2}$



    $displaystyle fleft(frac{x+z}{2}right) = lambda_1 f(z) + (1-lambda_1)fleft(frac{x+y+z}{3}right)$



    $displaystyle fleft(frac{y+z}{2}right) = lambda_2 f(z) + (1-lambda_2)fleft(frac{x+y+z}{3}right)$



    Adding the three inequalities gives the Popoviciu's inequality. The second case where $displaystyle y > frac{x+y+z}{3}$ we simply interchange the roles of $x$ and $z$ and the rest is similar.






    share|cite|improve this answer









    $endgroup$


















      3












      $begingroup$

      A proof of Popoviciu's inequality follows from Jensen's Inequality for convex functions. Wlog, assume $x le y le z$ there are two possibilities. Either $displaystyle y < frac{x+y+z}{3}$ or the reverse.



      In the first case, $displaystyle frac{x+y+z}{3} < frac{x+z}{2}< z$ and $displaystyle frac{x+y+z}{3} < frac{y+z}{2} < z$, thus, there are positive real numbers $lambda_1,lambda_2 >0$ such that,



      $displaystyle frac{x+z}{2} = lambda_1 z + (1-lambda_1)left(frac{x+y+z}{3}right)$ and $displaystyle frac{y+z}{2} = lambda_2 z + (1-lambda_2)left(frac{x+y+z}{3}right)$



      It is easy to verify that $lambda_1 + lambda_2 = dfrac{1}{2}$.



      Now, using Jensen's inequality one has:



      $displaystyle fleft(frac{x+y}{2}right) le frac{f(x)+f(y)}{2}$



      $displaystyle fleft(frac{x+z}{2}right) = lambda_1 f(z) + (1-lambda_1)fleft(frac{x+y+z}{3}right)$



      $displaystyle fleft(frac{y+z}{2}right) = lambda_2 f(z) + (1-lambda_2)fleft(frac{x+y+z}{3}right)$



      Adding the three inequalities gives the Popoviciu's inequality. The second case where $displaystyle y > frac{x+y+z}{3}$ we simply interchange the roles of $x$ and $z$ and the rest is similar.






      share|cite|improve this answer









      $endgroup$
















        3












        3








        3





        $begingroup$

        A proof of Popoviciu's inequality follows from Jensen's Inequality for convex functions. Wlog, assume $x le y le z$ there are two possibilities. Either $displaystyle y < frac{x+y+z}{3}$ or the reverse.



        In the first case, $displaystyle frac{x+y+z}{3} < frac{x+z}{2}< z$ and $displaystyle frac{x+y+z}{3} < frac{y+z}{2} < z$, thus, there are positive real numbers $lambda_1,lambda_2 >0$ such that,



        $displaystyle frac{x+z}{2} = lambda_1 z + (1-lambda_1)left(frac{x+y+z}{3}right)$ and $displaystyle frac{y+z}{2} = lambda_2 z + (1-lambda_2)left(frac{x+y+z}{3}right)$



        It is easy to verify that $lambda_1 + lambda_2 = dfrac{1}{2}$.



        Now, using Jensen's inequality one has:



        $displaystyle fleft(frac{x+y}{2}right) le frac{f(x)+f(y)}{2}$



        $displaystyle fleft(frac{x+z}{2}right) = lambda_1 f(z) + (1-lambda_1)fleft(frac{x+y+z}{3}right)$



        $displaystyle fleft(frac{y+z}{2}right) = lambda_2 f(z) + (1-lambda_2)fleft(frac{x+y+z}{3}right)$



        Adding the three inequalities gives the Popoviciu's inequality. The second case where $displaystyle y > frac{x+y+z}{3}$ we simply interchange the roles of $x$ and $z$ and the rest is similar.






        share|cite|improve this answer









        $endgroup$



        A proof of Popoviciu's inequality follows from Jensen's Inequality for convex functions. Wlog, assume $x le y le z$ there are two possibilities. Either $displaystyle y < frac{x+y+z}{3}$ or the reverse.



        In the first case, $displaystyle frac{x+y+z}{3} < frac{x+z}{2}< z$ and $displaystyle frac{x+y+z}{3} < frac{y+z}{2} < z$, thus, there are positive real numbers $lambda_1,lambda_2 >0$ such that,



        $displaystyle frac{x+z}{2} = lambda_1 z + (1-lambda_1)left(frac{x+y+z}{3}right)$ and $displaystyle frac{y+z}{2} = lambda_2 z + (1-lambda_2)left(frac{x+y+z}{3}right)$



        It is easy to verify that $lambda_1 + lambda_2 = dfrac{1}{2}$.



        Now, using Jensen's inequality one has:



        $displaystyle fleft(frac{x+y}{2}right) le frac{f(x)+f(y)}{2}$



        $displaystyle fleft(frac{x+z}{2}right) = lambda_1 f(z) + (1-lambda_1)fleft(frac{x+y+z}{3}right)$



        $displaystyle fleft(frac{y+z}{2}right) = lambda_2 f(z) + (1-lambda_2)fleft(frac{x+y+z}{3}right)$



        Adding the three inequalities gives the Popoviciu's inequality. The second case where $displaystyle y > frac{x+y+z}{3}$ we simply interchange the roles of $x$ and $z$ and the rest is similar.







        share|cite|improve this answer












        share|cite|improve this answer



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        answered Dec 19 '18 at 14:41









        r9mr9m

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