Ta có :

$$\frac{ab}{3+c^2}=\frac{ab}{(c^2+a^2)+(c^2+b^2)}\le \frac{1}{4}.\frac{(a+b{)}^2}{(c^2+a^2)+(c^2+b^2)}\le \frac{1}{4}(\frac{a^2}{c^2+a^2}+\frac{b^2}{c^2+b^2})\le \frac{1}{4}(\frac{a^2}{c^2+a^2}+\frac{b}{2c})$$

Hoàn toàn tương tự :

$$\frac{bc}{3+a^2}\le \frac{1}{4}(\frac{c^2}{c^2+a^2}+\frac{b}{2a})$$

Suy ra :

$$\frac{ab}{3+c^2}+\frac{bc}{3+a^2}\le \frac{1}{4}(1+\frac{1}{2}(\frac{b}{a}+\frac{b}{c}))=\frac{1}{4}+\frac{1}{8}(\frac{b}{a}+\frac{b}{c})$$

Và :

$$24A\le 6+3(\frac{b}{a}+\frac{b}{c})-(\frac{b^3}{a^3}+\frac{c^3}{a^3})$$

Dễ dàng chứng minh được :

$$3x-x^3\le 2(x>0)$$

Suy ra :

$$24A\le 6+(\frac{3b}{a}-\frac{b^3}{a^3})+(\frac{3b}{c}-\frac{b^3}{c^3})\le 6+2+2=10\iff A\le \frac{5}{12}$$

Kết luận :

$$A_{max}=\frac{5}{12}\iff a=b=c=1$$

Ta có :

$$(a^2+1)(b^2+1)=(a+b{)}^2+(ab-1{)}^2$$

$$2(c^2+1)=(c+1{)}^2+(c-1{)}^2$$

Áp dụng BĐT Cauchy-Schwarz :

$$\sqrt{2(a^2+1)(b^2+1)(c^2+1)}=\sqrt{\left[\right(a+b{)}^2+(ab-1{)}^2].\left[\right(c+1{)}^2+(c-1{)}^2]}\ge (a+b)(c+1)+(1-ab)(c-1)=(a+b+c+ab+bc+ca+abc+1)-2(1+abc)=(a+1)(b+1)(c+1)-2(1+abc)$$

Từ đó :

$$\frac{\sqrt{2(a^2+1)(b^2+1)(c^2+1)}}{(a+1)(b+1)}=\frac{(a+1)(b+1)(c+1)-2(1+abc)}{(a+1)(b+1)}=c+1-\frac{2(1+abc)}{(a+1)(b+1)}$$

Theo giả thiết :

$$\frac{12}{4ab+(a+b)(c+3)}=\frac{12}{3(a+b+ab)+(ab+bc+ca)}\ge \frac{12}{3(ab+a+b+1)}=\frac{4}{(a+1)(b+1)}$$

Từ đó mà :

$$P\ge \frac{2(1-abc)}{(a+1)(b+1)}+c+1+\frac{1}{2c^2}$$

Chú ý rằng :

$$3\ge ab+bc+ca\ge 3\sqrt[3]{3abc}\Rightarrow abc\le 1$$

$$c+\frac{1}{2c^2}=\frac{c}{2}+\frac{c}{2}+\frac{1}{2c^2}\ge 3\sqrt[3]{3\frac{1}{8}}=\frac{3}{2}$$

Như vậy ta được :

$$P\ge \frac{5}{2}$$

$$MinP=\frac{5}{2}\iff a=b=c=1$$