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Math Help - Inequality Proof

  1. #1
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    Inequality Proof

    Let a, b, and c be positive real numbers. Prove that

    \frac{a^3}{b+c}+\frac{b^3}{c+a}+\frac{c^3}{a+b} \geq \frac{a^2+b^2+c^2}{2}

    Thanks in advance!
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  2. #2
    Super Member PaulRS's Avatar
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    By the symmetry we may assume, WLOG, that a\geq{b}\geq{c}>0

    Then \frac{a}{b+c}\geq{\frac{b}{a+c}}\geq{\frac{c}{a+b}  }>0

    So, by Chebyshev's inequality we have:
    \frac{a^3}{b+c}+\frac{b^3}{c+a}+\frac{c^3}{a+b} \geq \frac{1}{3}\cdot{\left( \frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}\right)\c  dot{(a^2+b^2+c^2)}}

    And now, by Nesbitt's Inequality we have:  \frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}\geq{\tfr  ac{3}{2}}

    Thus \frac{a^3}{b+c}+\frac{b^3}{c+a}+\frac{c^3}{a+b} \geq \tfrac{1}{3}\cdot{\left(\tfrac{3}{2}\right)\cdot{(  a^2+b^2+c^2)}}=\frac{a^2+b^2+c^2}{2}
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  3. #3
    Super Member PaulRS's Avatar
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    Lemma. Suppose we have finite sequences: a_1,...,a_n;b_1,...,b_n;c_1,...,c_n of positive real numbers

    Then we have: <br />
\left( {\sum\limits_{k = 1}^n {a_k ^3 } } \right) \cdot \left( {\sum\limits_{k = 1}^n {b_k ^3 } } \right) \cdot \left( {\sum\limits_{k = 1}^n {c_k ^3 } } \right) \geqslant \left( {\sum\limits_{k = 1}^n {a_k  \cdot b_k  \cdot c_k } } \right)^3 <br />
( this is a particular case of Hölder's inequality, I'm proving it because I do not know whether you are familiar with it or not)

    Proof

    By AM-GM inequality: <br />
x^3  + y^3  + z^3  \geqslant 3 \cdot x \cdot y \cdot z<br />
( for x,y,z>0 )

    Thus: <br />
\tfrac{{a_k ^3 }}<br />
{{\sum\nolimits_{k = 1}^n {a_k ^3 } }} + \tfrac{{b_k ^3 }}<br />
{{\sum\nolimits_{k = 1}^n {b_k ^3 } }} + \tfrac{{c_k ^3 }}<br />
{{\sum\nolimits_{k = 1}^n {c_k ^3 } }} \geqslant 3 \cdot \tfrac{{a_k }}<br />
{{\sqrt[3]{{\sum\nolimits_{k = 1}^n {a_k ^3 } }}}} \cdot \tfrac{{b_k }}<br />
{{\sqrt[3]{{\sum\nolimits_{k = 1}^n {b_k ^3 } }}}} \cdot \tfrac{{c_k }}<br />
{{\sqrt[3]{{\sum\nolimits_{k = 1}^n {c_k ^3 } }}}}<br />

    ( set <br />
x = \tfrac{{a_k }}<br />
{{\sqrt[3]{{\sum\nolimits_{k = 1}^n {a_k ^3 } }}}}<br />
... above)

    Now sum the inequalities: <br />
\tfrac{{\sum\nolimits_{k = 1}^n {a_k ^3 } }}<br />
{{\sum\nolimits_{k = 1}^n {a_k ^3 } }} + \tfrac{{\sum\nolimits_{k = 1}^n {b_k ^3 } }}<br />
{{\sum\nolimits_{k = 1}^n {b_k ^3 } }} + \tfrac{{\sum\nolimits_{k = 1}^n {c_k ^3 } }}<br />
{{\sum\nolimits_{k = 1}^n {c_k ^3 } }} \geqslant <br />
<br />
3 \cdot \tfrac{{\sum\nolimits_{k = 1}^n {a_k b_k c_k } }}<br />
{{\left( {\sqrt[3]{{\sum\nolimits_{k = 1}^n {a_k ^3 } }}} \right) \cdot \left( {\sqrt[3]{{\sum\nolimits_{k = 1}^n {b_k ^3 } }}} \right) \cdot \left( {\sqrt[3]{{\sum\nolimits_{k = 1}^n {c_k ^3 } }}} \right)}}<br />

    Thus: <br />
3 \geqslant 3 \cdot \tfrac{{\sum\nolimits_{k = 1}^n {a_k b_k c_k } }}<br />
{{\left( {\sqrt[3]{{\sum\nolimits_{k = 1}^n {a_k ^3 } }}} \right) \cdot \left( {\sqrt[3]{{\sum\nolimits_{k = 1}^n {b_k ^3 } }}} \right) \cdot \left( {\sqrt[3]{{\sum\nolimits_{k = 1}^n {c_k ^3 } }}} \right)}}<br />
and the rest follows easily.


    Now let's go to your inequality.

    By the lemma we have: <br />
\left( {\tfrac{{a^3 }}<br />
{{b + c}} + \tfrac{{b^3 }}<br />
{{a + c}} + \tfrac{{c^3 }}<br />
{{a + b}}} \right) \cdot \left[ {a \cdot \left( {b + c} \right) + b \cdot \left( {a + c} \right) + c \cdot \left( {a + b} \right)} \right] \cdot \left( {a^2  + b^2  + c^2 } \right)<br />
<br />
 \geqslant \left( {a^2  + b^2  + c^2 } \right)^3 <br />

    Thus: <br />
\left( {\tfrac{{a^3 }}<br />
{{b + c}} + \tfrac{{b^3 }}<br />
{{a + c}} + \tfrac{{c^3 }}<br />
{{a + b}}} \right) \cdot \left[ {a \cdot \left( {b + c} \right) + b \cdot \left( {a + c} \right) + c \cdot \left( {a + b} \right)} \right] \geqslant \left( {a^2  + b^2  + c^2 } \right)^2 <br />

    Now note that: <br />
\left[ {a \cdot \left( {b + c} \right) + b \cdot \left( {a + c} \right) + c \cdot \left( {a + b} \right)} \right] = 2 \cdot \left( {ab + bc + ac} \right)<br />

    Thus we have: <br />
\left( {\tfrac{{a^3 }}<br />
{{b + c}} + \tfrac{{b^3 }}<br />
{{a + c}} + \tfrac{{c^3 }}<br />
{{a + b}}} \right) \cdot  \geqslant \tfrac{1}<br />
{2} \cdot \left( {a^2  + b^2  + c^2 } \right) \cdot \left( {\tfrac{{a^2  + b^2  + c^2 }}<br />
{{ab + bc + ac}}} \right)<br />

    But the following inequality holds: <br />
\tfrac{{a^2  + b^2  + c^2 }}<br />
{{ab + bc + ac}} \geqslant 1<br />
(*)

    Therefore: <br />
\left( {\tfrac{{a^3 }}<br />
{{b + c}} + \tfrac{{b^3 }}<br />
{{a + c}} + \tfrac{{c^3 }}<br />
{{a + b}}} \right) \cdot  \geqslant \tfrac{1}<br />
{2} \cdot \left( {a^2  + b^2  + c^2 } \right) \cdot \left( {\tfrac{{a^2  + b^2  + c^2 }}<br />
{{ab + bc + ac}}} \right) \geqslant \tfrac{1}<br />
{2} \cdot \left( {a^2  + b^2  + c^2 } \right)\square <br />

    To prove (*) note that: <br />
\left( {a - b} \right)^2  \geqslant 0 \Leftrightarrow a^2  + b^2  \geqslant 2ab<br />

    Similarly: <br />
\left[ \begin{gathered}<br />
  a^2  + b^2  \geqslant 2ab \hfill \\<br />
  a^2  + c^2  \geqslant 2ac \hfill \\<br />
  b^2  + c^2  \geqslant 2bc \hfill \\ <br />
\end{gathered}  \right.<br />
and sum those inequalities.
    Last edited by PaulRS; December 22nd 2008 at 11:07 AM. Reason: I'd written positive integers instead of positive real numbers (statement of the lemma)
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