1. ## Inequality Proof

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

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

2. By the symmetry we may assume, WLOG, that $\displaystyle a\geq{b}\geq{c}>0$

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

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

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

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

3. Lemma. Suppose we have finite sequences: $\displaystyle a_1,...,a_n;b_1,...,b_n;c_1,...,c_n$ of positive real numbers

Then we have: $\displaystyle \left( {\sum\limits_{k = 1}^n {a_k ^3 } } \right) \cdot \left( {\sum\limits_{k = 1}^n {b_k ^3 } } \right) \cdot \left( {\sum\limits_{k = 1}^n {c_k ^3 } } \right) \geqslant \left( {\sum\limits_{k = 1}^n {a_k \cdot b_k \cdot c_k } } \right)^3$ ( 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: $\displaystyle x^3 + y^3 + z^3 \geqslant 3 \cdot x \cdot y \cdot z$ ( for $\displaystyle x,y,z>0$ )

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

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

Now sum the inequalities: $\displaystyle \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 $$\displaystyle 3 \cdot \tfrac{{\sum\nolimits_{k = 1}^n {a_k b_k c_k } }} {{\left( {\sqrt[3]{{\sum\nolimits_{k = 1}^n {a_k ^3 } }}} \right) \cdot \left( {\sqrt[3]{{\sum\nolimits_{k = 1}^n {b_k ^3 } }}} \right) \cdot \left( {\sqrt[3]{{\sum\nolimits_{k = 1}^n {c_k ^3 } }}} \right)}} Thus: \displaystyle 3 \geqslant 3 \cdot \tfrac{{\sum\nolimits_{k = 1}^n {a_k b_k c_k } }} {{\left( {\sqrt[3]{{\sum\nolimits_{k = 1}^n {a_k ^3 } }}} \right) \cdot \left( {\sqrt[3]{{\sum\nolimits_{k = 1}^n {b_k ^3 } }}} \right) \cdot \left( {\sqrt[3]{{\sum\nolimits_{k = 1}^n {c_k ^3 } }}} \right)}} and the rest follows easily. Now let's go to your inequality. By the lemma we have: \displaystyle \left( {\tfrac{{a^3 }} {{b + c}} + \tfrac{{b^3 }} {{a + c}} + \tfrac{{c^3 }} {{a + b}}} \right) \cdot \left[ {a \cdot \left( {b + c} \right) + b \cdot \left( {a + c} \right) + c \cdot \left( {a + b} \right)} \right] \cdot \left( {a^2 + b^2 + c^2 } \right)$$\displaystyle \geqslant \left( {a^2 + b^2 + c^2 } \right)^3$

Thus: $\displaystyle \left( {\tfrac{{a^3 }} {{b + c}} + \tfrac{{b^3 }} {{a + c}} + \tfrac{{c^3 }} {{a + b}}} \right) \cdot \left[ {a \cdot \left( {b + c} \right) + b \cdot \left( {a + c} \right) + c \cdot \left( {a + b} \right)} \right] \geqslant \left( {a^2 + b^2 + c^2 } \right)^2$

Now note that: $\displaystyle \left[ {a \cdot \left( {b + c} \right) + b \cdot \left( {a + c} \right) + c \cdot \left( {a + b} \right)} \right] = 2 \cdot \left( {ab + bc + ac} \right)$

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

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

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

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

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