Cho $a,b,c$ là các số thực không âm thỏa mãn điều kiện $ab+bc+ca=3$ và $c\le a.$

Tìm giá trị nhỏ nhát của biểu thức $P=\frac{1}{{\left(a+1\right)}^2}+\frac{2}{{\left(b+1\right)}^2}+\frac{3}{{\left(c+1\right)}^2}.$

Cách 1: Theo đề bài $ab+bc+ca=3.$ Áp dụng bất đẳng thức Cauchy ta có

$$\begin{gathered} 3=ab+bc+ac\ge 3\sqrt[3]{a^2{b}^2{c}^2}\Rightarrow abc\le 1,\left(1\right) \\ {\left(a+b+c\right)}^2\ge 3\left(ab+bc+ac\right)=9\Rightarrow a+b+c\ge 3,\left(2\right) \\ Từ \left(1\right) và \left(2\right) \Rightarrow a+b+c\ge 3abc. \\ Đặt x=\frac{1}{a+1};y=\frac{1}{b+1};z=\frac{1}{c+1}\left(\Rightarrow x,y,z>0;z\ge x\right) \\ \Rightarrow P=x^2+2y^2+3z^2=x^2+z^2+2y^2+2z^2\ge 2\left(x^2+y^2+z^2\right) \\ \Rightarrow P\ge 2\left(x^2+y^2+z^2\right)\ge 2\left(xy+yz+xz\right).(*) \end{gathered}$$

$$\begin{gathered} Ta tìm giá trị nhỏ nhất của xy+yz+xz . \\ xy+yz+xz=\frac{1}{\left(a+1\right)\left(b+1\right)}+\frac{1}{\left(c+1\right)\left(b+1\right)}+\frac{1}{\left(a+1\right)\left(c+1\right)} \\ \iff xy+yz+xz=\frac{a+b+c+3}{\left(a+1\right)\left(b+1\right)\left(c+1\right)}=\frac{a+b+c+3}{abc+a+b+c+4} \\ \iff xy+yz+xz=\frac{a+b+c+3}{abc+a+b+c+4}=\frac{3\left(a+b+c+3\right)}{3abc+3\left(a+b+c\right)+12} \\ \iff xy+yz+xz=\frac{3\left(a+b+c+3\right)}{3abc+3\left(a+b+c\right)+12}\ge \frac{3\left(a+b+c+3\right)}{\left(a+b+c\right)+3\left(a+b+c\right)+12}=\frac{3}{4} \\ \Rightarrow P\ge 2.\frac{3}{4}=\frac{3}{2}. \end{gathered}$$

Dấu bằng xảy ra khi $x=y=z\Rightarrow a=b=c=1.$

Vậy giá trị nhỏ nhất của $P=\frac{3}{2}.$

Cách 2: Vì $Vì a\ge c\Rightarrow P=\frac{1}{{\left(a+1\right)}^2}+\frac{2}{{\left(b+1\right)}^2}+\frac{3}{{\left(c+1\right)}^2}\ge \frac{1}{{\left(a+1\right)}^2}+\frac{2}{{\left(b+1\right)}^2}+\frac{2}{{\left(c+1\right)}^2}+\frac{1}{{\left(a+1\right)}^2}$

$\Rightarrow P\ge \frac{2}{{\left(a+1\right)}^2}+\frac{2}{{\left(b+1\right)}^2}+\frac{2}{{\left(c+1\right)}^2}$

Ta chứng minh đẳng thức với x, y không âm.

$\frac{1}{{\left(x+1\right)}^2}+\frac{1}{{\left(y+1\right)}^2}\ge \frac{1}{1+xy}$

$$\begin{gathered} \iff \left(1+xy\right)\left(x^2+y^2+2x+2y+2\right)-{\left(xy+x+y+1\right)}^2\ge 0 \\ \iff \left(1+xy\right)\left(x^2+y^2-2xy+2xy+2x+2y+2\right)-{\left(xy+x+y+1\right)}^2\ge 0 \\ \iff \left(1+xy\right){\left(x-y\right)}^2+2\left(1+xy\right)\left(xy+x+y+1\right)-{\left(xy+x+y+1\right)}^2\ge 0 \\ \iff \left(1+xy\right){\left(x-y\right)}^2+\left(xy-x-y+1\right)\left(xy+x+y+1\right)\ge 0 \\ \iff xy{\left(x-y\right)}^2+{\left(x-y\right)}^2+{\left(xy+1\right)}^2-{\left(x+y\right)}^2\ge 0 \\ \iff xy{\left(x-y\right)}^2+{\left(xy-1\right)}^2\ge 0. \end{gathered}$$

Luôn đúng, dấu "=" xảy ra khi x=y=z=1.

$\Rightarrow P=\frac{1}{{\left(a+1\right)}^2}+\frac{2}{{\left(b+1\right)}^2}+\frac{3}{{\left(c+1\right)}^2}\ge \frac{1}{{\left(a+1\right)}^2}+\frac{2}{{\left(b+1\right)}^2}+\frac{2}{{\left(c+1\right)}^2}+\frac{1}{{\left(a+1\right)}^2}$

$\Rightarrow P\ge \frac{1}{{\left(a+1\right)}^2}+\frac{1}{{\left(b+1\right)}^2}+\frac{1}{{\left(b+1\right)}^2}+\frac{1}{{\left(c+1\right)}^2}+\frac{1}{{\left(a+1\right)}^2}\ge \frac{1}{1+ab}+\frac{1}{1+bc}+\frac{1}{1+ac}.$

Áp dụng BĐT Cauchy cho 3 số không âm ta có

$$\begin{gathered} \left(x+y+z\right)\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)\ge 9\Rightarrow \frac{1}{x}+\frac{1}{y}+\frac{1}{z}\ge \frac{9}{x+y+z} \\ \Rightarrow P\ge \frac{1}{1+ab}+\frac{1}{1+bc}+\frac{1}{1+ac}\ge \frac{9}{3+ab+bc+ac}=\frac{9}{6}=\frac{3}{2}. \end{gathered}$$

Vậy GTNN của $P=\frac{3}{2}$ khi $a=b=c=1.$