a) Chứng minh với mọi số thực a,...

- Hướng dẫn giải

Phương pháp giải:

Giải chi tiết:

a) Chứng minh với mọi số thực a, b, c ta có \(ab+bc+ca\le \frac{{{\left( a+b+c \right)}^{2}}}{3}\)

\(\begin{align}  & \,\,\,\,\,ab+bc+ca\le \frac{{{\left( a+b+c \right)}^{2}}}{3} \\  & \Leftrightarrow 3\left( ab+bc+ca \right)\le {{\left( a+b+c \right)}^{2}} \\  & \Leftrightarrow 3\left( ab+bc+ca \right)\le {{a}^{2}}+{{b}^{2}}+{{c}^{2}}+2\left( ab+bc+ca \right) \\  & \Leftrightarrow ab+bc+ca\le {{a}^{2}}+{{b}^{2}}+{{c}^{2}} \\  & \Leftrightarrow {{a}^{2}}+{{b}^{2}}+{{c}^{2}}-\left( ab+bc+ca \right)\ge 0 \\  & \Leftrightarrow 2\left( {{a}^{2}}+{{b}^{2}}+{{c}^{2}} \right)-2\left( ab+bc+ca \right)\ge 0 \\  & \Leftrightarrow \left( {{a}^{2}}-2ab+{{b}^{2}} \right)+\left( {{b}^{2}}-2bc+{{c}^{2}} \right)+\left( {{c}^{2}}-2ca+{{a}^{2}} \right)\ge 0 \\  & \Leftrightarrow {{\left( a-b \right)}^{2}}+{{\left( b-c \right)}^{2}}+{{\left( c-a \right)}^{2}}\ge 0\,\,\left( luon\,\,dung \right) \\ \end{align}\)

b) Cho ba số dương x, y, z thỏa mãn điều kiện \(x+y+z=\frac{3}{4}\)  Chứng minh

\(6\left( {{x}^{2}}+{{y}^{2}}+{{z}^{2}} \right)+10\left( xy+yz+xz \right)+2\left( \frac{1}{2x+y+z}+\frac{1}{x+2y+z}+\frac{1}{x+y+2z} \right)\ge 9\)

Đẳng thức xảy ra khi nào?

\(\begin{align}  & 6\left( {{x}^{2}}+{{y}^{2}}+{{z}^{2}} \right)+10\left( xy+yz+xz \right)+2\left( \frac{1}{2x+y+z}+\frac{1}{x+2y+z}+\frac{1}{x+y+2z} \right)\ge 9 \\  & \Leftrightarrow 6\left( {{x}^{2}}+{{y}^{2}}+{{z}^{2}}+2\left( xy+yz+xz \right) \right)-12\left( xy+yz+xz \right)+10\left( xy+yz+xz \right) \\  & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,+2\left( \frac{1}{2x+y+z}+\frac{1}{x+2y+z}+\frac{1}{x+y+2z} \right)\ge 9 \\  & \Leftrightarrow 6{{\left( x+y+z \right)}^{2}}-2\left( xy+yz+xz \right)+2\left( \frac{1}{2x+y+z}+\frac{1}{x+2y+z}+\frac{1}{x+y+2z} \right)\ge 9 \\  & \Leftrightarrow 6.{{\left( \frac{3}{4} \right)}^{2}}-2\left( xy+yz+xz \right)+2\left( \frac{1}{2x+y+z}+\frac{1}{x+2y+z}+\frac{1}{x+y+2z} \right)\ge 9 \\  & \Leftrightarrow \frac{27}{8}-2\left( xy+yz+xz \right)+2\left( \frac{1}{2x+y+z}+\frac{1}{x+2y+z}+\frac{1}{x+y+2z} \right)\ge 9\,\,\,\left( * \right) \\ \end{align}\)

Áp dụng BĐT ở ý a) ta có:

\(xy+yz+zx\le \frac{{{\left( x+y+z \right)}^{2}}}{3}=\frac{{{\left( \frac{3}{4} \right)}^{2}}}{3}=\frac{3}{16}\Rightarrow -2\left( xy+yz+zx \right)\ge \frac{-3}{8}\)

Áp dụng BĐT phụ: \(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\ge \frac{9}{a+b+c}\,\,\left( a,b,c\ge 0 \right)\)

Chứng minh:

Ta có: \(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\ge \frac{9}{a+b+c}\Leftrightarrow \left( \frac{1}{a}+\frac{1}{b}+\frac{1}{c} \right)\left( a+b+c \right)\ge 9\)

Áp dụng BĐT Cô-si ta có: \(\left\{ \begin{align}  & \frac{1}{a}+\frac{1}{b}+\frac{1}{c}\ge 3\sqrt{\frac{1}{abc}} \\  & a+b+c\ge 3\sqrt{abc} \\ \end{align} \right.\Rightarrow \left( \frac{1}{a}+\frac{1}{b}+\frac{1}{c} \right)\left( a+b+c \right)\ge 3\sqrt{\frac{1}{abc}}.3\sqrt{abc}=9\,\,\left( dpcm \right)\)

Ta có:   

\(\frac{1}{2x+y+z}+\frac{1}{x+2y+z}+\frac{1}{x+y+2z}\ge \frac{9}{2x+y+z+x+2y+z+x+y+2z}=\frac{9}{4\left( x+y+z \right)}=\frac{9}{4.\frac{3}{4}}=3\)

\(\Rightarrow V{{T}_{\left( * \right)}}\ge \frac{27}{8}-\frac{3}{8}+2.3\ge 9\,\,\left( dpcm \right)\)