Ta có : \(ab+bc+ca=2abc\)

\(\Leftrightarrow\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=2\)

Đặt \(\hept{\begin{cases}x=\frac{1}{a}\\y=\frac{1}{b}\\z=\frac{1}{c}\end{cases}}\)

\(\Rightarrow\hept{\begin{cases}x+y+z=2\\P=\frac{x^3}{\left(2-x\right)^2}+\frac{y^3}{\left(2-y\right)^3}+\frac{z^3}{\left(2-z\right)^2}\end{cases}}\)

Áp dụng bất đẳng thức Cauchy - Schwarz 

\(\Rightarrow\frac{x^3}{\left(2-x\right)^2}+\frac{2-x}{8}+\frac{2-x}{8}\ge3\sqrt[3]{\frac{x^3}{64}}=\frac{3x}{4}\)

Tương tự ta có :

\(\hept{\begin{cases}\frac{y^3}{\left(2-y\right)^2}+\frac{2-y}{8}+\frac{2-y}{8}\ge\frac{3y}{4}\\\frac{z^3}{\left(2-z\right)^2}+\frac{2-z}{8}+\frac{2-z}{8}\ge\frac{3z}{8}\end{cases}}\)

\(\Rightarrow P+\frac{12-2\left(x+y+z\right)}{8}\ge\frac{3}{4}\left(x+y+z\right)\)

\(\Rightarrow P\ge\frac{1}{12}\)

Dấu " = " xảy ra khi \(x=y=z=\frac{2}{3}\)

Ta có : \(ab+bc+ca=2abc\)

\(\Leftrightarrow\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=2\)

Đặt \(\hept{\begin{cases}x=\frac{1}{a}\\y=\frac{1}{b}\\z=\frac{1}{c}\end{cases}}\)

\(\Rightarrow\hept{\begin{cases}x+y+z=2\\P=\frac{x^3}{\left(2-x\right)^2}+\frac{y^3}{\left(2-y\right)^3}+\frac{z^3}{\left(2-z^2\right)}\end{cases}}\)

Áp dụng bất đẳng thức Cauchy - Schwarz 

\(\Rightarrow\frac{x^3}{\left(2-x\right)^2}+\frac{2-x}{8}+\frac{2-x}{8}\ge3\sqrt[3]{\frac{x^3}{64}}=\frac{3x}{4}\)

Tương tự ta có : \(\hept{\begin{cases}\frac{y^3}{\left(2-y\right)^2}+\frac{2-y}{8}+\frac{2-y}{8}\ge\frac{3y}{4}\\\frac{z^3}{\left(2-z\right)^2}+\frac{2-z}{8}+\frac{2-z}{8}\ge\frac{3z}{8}\end{cases}}\)

\(\Rightarrow P+\frac{12-2\left(x+y+z\right)}{8}\ge\frac{3}{4}\left(x+y+z\right)\)

\(\Rightarrow P\ge\frac{1}{2}\)

Dấu "=" xảy ra khi \(x=y=z=\frac{2}{3}\)

1/\(=4a^2+4b^2+c^2+8ab-4bc-4ca+4b^2+4c^2+a^2+8bc-4ca-4ab+4a^2+4c^2+b^2+8ca-4bc-4ab=\)

\(=9a^2+9b^2+9c^2=9\left(a^2+b^2+c^2\right)\)

2/

Ta có

\(\left(a+b+c\right)^2=a^2+b^2+c^2+2\left(ab+bc+ca\right)\ge0\)

\(\Leftrightarrow a^2+b^2+c^2\ge-2\left(ab+bc+ca\right)=2\)

\(\Rightarrow P=9\left(a^2+b^2+c^2\right)\ge18\)

\(\Rightarrow P_{min}=18\)

Đặt \(x=\frac{1}{a}, y=\frac{1}{b}, z=\frac{1}{c}, \Rightarrow x+y+z=2\)

Suy ra    \(\frac{1}{a\left(2a-1\right)^2}+\frac{1}{b\left(2b-1\right)^2}+\frac{1}{c\left(2c-1\right)^2}=\frac{x^3}{\left(2-x\right)^2}+\frac{y^3}{\left(2-y\right)^2}+\frac{z^3}{\left(2-z\right)^2}\)

Ta có \(\frac{x^3}{\left(2-x\right)^2}+\frac{2-x}{8}+\frac{2-x}{8}\ge3\sqrt[3]{\frac{x^3}{\left(2-x\right)^2} .\frac{2-x}{8}.\frac{2-x}{8}}=\frac{3x}{4}.\)

\(\Rightarrow\frac{x^3}{\left(2-x\right)^2}\ge x-\frac{1}{2}\)\(\Rightarrow\frac{x^3}{\left(2-x\right)^2}+\frac{y^3}{\left(2-y\right)^2}+\frac{z^3}{\left(2-z\right)^2}\ge x+y+z-\frac{3}{2}=2-\frac{3}{2}=\frac{1}{2}\)

dấu "=" xảy ra khi \(x=y=z=\frac{2}{3}\)hay \(a=b=c=\frac{3}{2}\)

Đặt \(\frac{1}{a}=x>0;\frac{1}{b}=y>0;\frac{1}{c}=z>0\)

Từ giả thiết ta có: \(7\left(x^2+y^2+z^2\right)=6\left(xy+yz+zx\right)+2015\le6\left(x^2+y^2+z^2\right)+2015\)

\(\Leftrightarrow x^2+y^2+z^2\le2015\)

Ta có: \(\frac{1}{\sqrt{3\left(2a^2+b^2\right)}}=\frac{1}{\sqrt{\left(4a^2+b^2\right)+\left(2a^2+2b^2\right)}}\le\frac{1}{\sqrt{4a^2+b^2+4ab}}=\frac{1}{2a+b}=\frac{1}{a+a+b}\le\frac{1}{9}\left(\frac{2}{a}+\frac{1}{b}\right)=\frac{1}{9}\left(2x+y\right)\)

Tương tự thì: \(\frac{1}{\sqrt{3\left(2b^2+c^2\right)}}\le\frac{1}{9}\left(2y+z\right)\)  và \(\frac{1}{\sqrt{3\left(2c^2+a^2\right)}}\le\frac{1}{9}\left(2z+x\right)\)

Cộng từng vế 3 BĐT trên ta có: \(\frac{1}{\sqrt{3\left(2a^2+b^2\right)}}+\frac{1}{\sqrt{3\left(2b^2+c^2\right)}}+\frac{1}{\sqrt{3\left(2c^2+a^2\right)}}\le\frac{x+y+z}{3}\le\frac{\sqrt{3\left(x^2+y^2+z^2\right)}}{3}\le\sqrt{\frac{2015}{3}}\)

Vậy max \(P=\sqrt{\frac{2015}{3}}\)  , đạt được khi \(a=b=c=\sqrt{\frac{3}{2015}}\)

\(a^2b^2c^2+\left(a+1\right)\left(1+b\right)\left(1+c\right)\ge a+b+c+ab+bc+ca+3\)

\(\Leftrightarrow\left(abc\right)^2+abc-2\ge0\Leftrightarrow\left(abc+2\right)\left(abc-1\right)\ge0\Leftrightarrow abc\ge1\)

Áp dụng BĐT Cosi ta có:

\(\frac{a^3}{\left(b+2c\right)\left(2c+3a\right)}+\frac{b+2c}{45}+\frac{2c+3a}{75}\ge3\sqrt[3]{\frac{a^3}{\left(b+2c\right)\left(2c+3b\right)}\cdot\frac{b+2c}{45}\cdot\frac{2c+3a}{75}}=\frac{a}{5}\left(1\right)\)

Tương tự ta có: \(\hept{\begin{cases}\frac{b^3}{\left(c+2a\right)\left(2a+3b\right)}+\frac{c+2a}{45}+\frac{2a+3b}{75}\ge\frac{b}{5}\left(2\right)\\\frac{c^3}{\left(a+2b\right)\left(2b+3c\right)}+\frac{a+2b}{45}+\frac{2b+3c}{75}\ge\frac{c}{5}\left(3\right)\end{cases}}\)

Từ (1)(2)(3) ta có:

\(P+\frac{2\left(a+b+c\right)}{15}\ge\frac{a+b+c}{5}\Leftrightarrow P\ge\frac{1}{15}\left(a+b+c\right)\)

Mà \(a+b+c\ge3\sqrt[3]{abc}\Rightarrow S\ge\frac{1}{5}\)

Dấu "=" xảy ra <=> a=b=c=1

CHÚC BAN HỌC GIỎI

Phạm Thị Diệu Huyền

Nguyễn Việt Lâm

Phạm Minh Quang

Trần Thanh Phương

Akai Haruma