homogenous and symetrically, so we can say $1=b\gec\ge0.$ We see already the ineq is equivalent with $\left(b-c\right)^{4022}\left(b+c\right)^{2011}\leq\left(b^{2011}-c^{2011}\right)^2\left(b^{2011}+c^{2011}\right).$ And $\left(b-c\right)^{2011}\le\left(b^{2011}-c^{2011}\right)$ with equality if $b=c$ or $c=0.$ The rest is equivalent with $\sqrt{b^2-c^2}\le\sqrt[2011]{b^{2011}-c^{2011}}.$ Now we let $f(i)=\sqrt[i]{1-x^i}$ and look for convex/concavity, would be a way I think. Otherwise $(1-c^2)^{2011}\le(1-c^{2011})^2=1-2c^{2011}+c^{4022}$ equivalent with $1-2c^2\le1-2c^{2011}$ which is true.

Without loss of generality, we assume that $\displaystyle {b \ge c.}$ The demonstrable inequality written equivalent: $\displaystyle {\left ({{b ^ {2011}} - {c ^ {2011}}} \right) \left ({{b ^ {2011}} + {c ^ {2011}}} \right) \left ({{b ^ {2011}} - {c ^ {2011}}} \right) \ge {\left ({b - c} \right) ^ {2011}} {\left ({b + c} \right) ^ {2011}} {\left ({b - c} \right) ^ {2011}} \Leftrightarrow}$ $\displaystyle {\Leftrightarrow \left ({{b ^ {4022}} - {c ^ {4022}}} \right) \left ({{b ^ {2011}} - {c ^ {2011}}} \right ) \ge {\left ({{b ^ 2} - {c ^ 2}} \right) ^ {2011}} {\left ({b - c} \right) ^ {2011}}} \bf \color {red} \left (1 \right).$ Since the type of Newton's binomial that we $\displaystyle {{b ^ {2011}} = {\left [{\left ({b - c} \right) + c} \right] ^ {2011}} = {\left ({b - c} \right ) ^ {2011}} + \cdots + {c ^ {2011}} \ge {\left ({b - c} \right) ^ {2011}} + {c ^ {2011}},}$ so $\displaystyle {{b ^ {2011}} - {c ^ {2011}} \ge {\left ({b - c} \right) ^ {2011}}} \bf \color {red} \left (2 \right).$ Putting the relationship $\bf \color {red} \left (2 \right) $ where $\displaystyle {b}$ and $\displaystyle {c}$ the $\displaystyle {b ^ 2}$ and $\displaystyle {c ^ 2}$ respectively, we have that: $\displaystyle {{b ^ {4022}} - {c ^ {4022}} \ge {\left ({{b ^ 2} - {c ^ 2}} \right) ^ {2011}}} \bf \color {red} \left (3 \right).$ The demonstrable relationship $\bf \color {red} \left (1 \right) $is obtained by multiplying $\bf \color {red} \left (2 \right)$ and $\bf \color {red} \left (3 \right)$ by members.