1. ## Inequality Proof

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

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

2. 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 $\frac{a^3}{b+c}+\frac{b^3}{c+a}+\frac{c^3}{a+b} \geq \tfrac{1}{3}\cdot{\left(\tfrac{3}{2}\right)\cdot{( a^2+b^2+c^2)}}=\frac{a^2+b^2+c^2}{2}$

3. 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: $
\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
$
( 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: $
x^3 + y^3 + z^3 \geqslant 3 \cdot x \cdot y \cdot z
$
( for $x,y,z>0$ )

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

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

Now sum the inequalities: $
\tfrac{{\sum\nolimits_{k = 1}^n {a_k ^3 } }}
{{\sum\nolimits_{k = 1}^n {a_k ^3 } }} + \tfrac{{\sum\nolimits_{k = 1}^n {b_k ^3 } }}
{{\sum\nolimits_{k = 1}^n {b_k ^3 } }} + \tfrac{{\sum\nolimits_{k = 1}^n {c_k ^3 } }}
{{\sum\nolimits_{k = 1}^n {c_k ^3 } }} \geqslant
$
$
3 \cdot \tfrac{{\sum\nolimits_{k = 1}^n {a_k b_k c_k } }}
{{\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)}}
$

Thus: $
3 \geqslant 3 \cdot \tfrac{{\sum\nolimits_{k = 1}^n {a_k b_k c_k } }}
{{\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)}}
$
and the rest follows easily.

Now let's go to your inequality.

By the lemma we have: $
\left( {\tfrac{{a^3 }}
{{b + c}} + \tfrac{{b^3 }}
{{a + c}} + \tfrac{{c^3 }}
{{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)
$
$
\geqslant \left( {a^2 + b^2 + c^2 } \right)^3
$

Thus: $
\left( {\tfrac{{a^3 }}
{{b + c}} + \tfrac{{b^3 }}
{{a + c}} + \tfrac{{c^3 }}
{{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
$

Now note that: $
\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)
$

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

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

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

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

Similarly: $
\left[ \begin{gathered}
a^2 + b^2 \geqslant 2ab \hfill \\
a^2 + c^2 \geqslant 2ac \hfill \\
b^2 + c^2 \geqslant 2bc \hfill \\
\end{gathered} \right.
$
and sum those inequalities.