неравенства - Доказать, что $%\dfrac{(bc)^2}{a+1}+\dfrac{(ac)^2}{b+1}+\dfrac{(ab)^2}{c+1}\le\dfrac{1}{36}$%

$$\dfrac {(bc)^2}{a}\cdot\dfrac {a}{1+a}+\dfrac {(ca)^2}{b}\cdot\dfrac {b}{1+b}+\dfrac {(ab )^2}{c}\dfrac {c}{1+c}\le\dfrac {b^2c^2+c^2a^2+a^2b^2}{1+\dfrac {b^2c^2+c^2a^2+a^2b^2 }{ \dfrac {b^2c^2}{a}+\dfrac {c^2a^2}{b}+\dfrac {a^2b^2}{c} }} \le \dfrac {1}{36}$$

$$\Leftrightarrow 36 (a^2b^2+b^2c^2+c^2a^2) \le 1+\dfrac {a^2b^2+b^2c^2+c^2a^2}{ \dfrac {a^2b^2}{c}+\dfrac {b^2c^2}{a}+\dfrac {c^2a^2}{b}} $$

$$\sqrt {\dfrac {a^2b^2}{c^2}+\dfrac{ b^2c^2}{a^2}+\dfrac {c^2a^2}{b^2}}\sqrt {a^2b^2+b^2c^2+c^2a^2}\ge \dfrac {a^2b^2}{c}+\dfrac {b^2c^2}{a}+\dfrac {c^2a^2}{b} $$

$$ 36 (a^2b^2+b^2c^2+c^2a^2) \le1+\sqrt {\dfrac {a^2b^2c^2 (a^2b^2+b^2c^2+c^2a^2)}{a^4b^4+b^4c^4+c^4a^4}} \le 1+\dfrac {a^2b^2+b^2c^2+c^2a^2}{ \dfrac {a^2b^2}{c}+\dfrac {b^2c^2}{a}+\dfrac {c^2a^2}{b}} $$

$$a^2=x\ , \ b^2=y\ , \ c^2=z$$

$$p=x+y+z=\dfrac {1}{3}\ , \ \dfrac {1}{36}\le xy+yz+zx=q\le\dfrac {1}{27}\ ,\ xyz=r$$

$$36q\le 1+\sqrt {\dfrac {rq}{q^2-2/3r}}$$

По н. ШУРА:

$$p^3+9r \ge4pq \Rightarrow r\ge \dfrac {4}{27}q-\dfrac {1}{243}$$

$$36q\le 1+\sqrt {\dfrac {q (36q-1)}{243q^2-24q+2/3}}$$

$$\dfrac {(1-27q)(36q-1)(972q^2-87q+2)}{729q^2-72q+2} \ge 0$$

Равенство при: $%\left (q=\dfrac {1}{36}:\ \ \ a=0, b=c=\dfrac {1}{\sqrt 6}\right)\ ,\ \left (q=\dfrac {1}{27}: \ \ a=b=c=\dfrac {1}{3}\right)$%