a) Cho hai số \(x>0,\,\,y>0\). Chứng minh rằ...

- Hướng dẫn giải

Phương pháp giải:

a) Biến đổi tương đương.

b) Áp dụng BĐT \(\frac{1}{x+y}\le \frac{1}{4}\left( \frac{1}{x}+\frac{1}{y} \right)\) và \(\frac{1}{a+b+c}\le \frac{1}{9}\left( \frac{1}{a}+\frac{1}{b}+\frac{1}{c} \right)\).

Giải chi tiết:

\(\begin{align}  & a)\,\,\frac{1}{x+y}\le \frac{1}{4}\left( \frac{1}{x}+\frac{1}{y} \right) \\  & \Leftrightarrow \frac{1}{x+y}\le \frac{x+y}{4xy} \\  & \Leftrightarrow {{\left( x+y \right)}^{2}}\ge 4xy\ \ \ \ \left( do\ \ x>0,\ y>0 \right) \\  & \Leftrightarrow {{x}^{2}}+{{y}^{2}}+2xy\ge 4xy \\  & \Leftrightarrow {{x}^{2}}-2xy+{{y}^{2}}\ge 0 \\  & \Leftrightarrow {{\left( x-y \right)}^{2}}\ge 0\,\,\left( luon\,\,dung \right) \\ \end{align}\)

b) Với ba số \(a,\ b,\ c>0,\)  ta chứng minh BĐT \(\frac{1}{a+b+c}\le \frac{1}{9}\left( \frac{1}{a}+\frac{1}{b}+\frac{1}{c} \right)\)

\(\begin{align}  & \Leftrightarrow \frac{1}{a+b+c}\le \frac{1}{9}\frac{ab+bc+ca}{abc} \\  & \Leftrightarrow 9abc\le \left( a+b+c \right)\left( ab+bc+ca \right) \\  & \Leftrightarrow 18abc\le \left( a+b+c \right)\left( 2ab+2bc+2ca \right)\,\,\,\left( * \right) \\ \end{align}\)

Ta có

\(\begin{align}  & {{\left( a+b+c \right)}^{2}}={{a}^{2}}+{{b}^{2}}+{{c}^{2}}+2ab+2bc+2ca \\  & \Rightarrow {{\left( a+b+c \right)}^{2}}\ge 2ab+2bc+2ca \\  & \Rightarrow 18abc\le {{\left( a+b+c \right)}^{3}} \\ \end{align}\)

Mà \({{\left( a+b+c \right)}^{3}}\overset{Cauchy}{\mathop{\ge }}\,{{\left( 3\sqrt[3]{abc} \right)}^{3}}=27abc\ge 18abc\)

Vậy \(\frac{1}{a+b+c}\le \frac{1}{9}\left( \frac{1}{a}+\frac{1}{b}+\frac{1}{c} \right)\)

Dấu bằng xảy ra \(\Leftrightarrow a=b=c\).

Áp dụng BĐT \(\frac{1}{x+y}\le \frac{1}{4}\left( \frac{1}{x}+\frac{1}{y} \right)\) và \(\frac{1}{a+b+c}\le \frac{1}{9}\left( \frac{1}{a}+\frac{1}{b}+\frac{1}{c} \right)\) ta có:

\(\begin{align}  & VT=\frac{1}{3a+2b+c}+\frac{1}{a+3b+2c}+\frac{1}{2a+b+3c} \\  & =\frac{1}{a+b+a+b+a+c}+\frac{1}{a+b+b+c+b+c}+\frac{1}{a+c+b+c+a+c} \\  & \le \frac{1}{9}\left( \frac{1}{a+b}+\frac{1}{a+b}+\frac{1}{a+c}+\frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{b+c}+\frac{1}{a+c}+\frac{1}{b+c}+\frac{1}{a+c} \right) \\  & =\frac{1}{3}\left( \frac{1}{a+b}+\frac{1}{a+c}+\frac{1}{b+c} \right) \\  & \le \frac{1}{3}.\frac{1}{4}\left( \frac{1}{a}+\frac{1}{b}+\frac{1}{a}+\frac{1}{c}+\frac{1}{b}+\frac{1}{c} \right) \\  & =\frac{1}{6}\left( \frac{1}{a}+\frac{1}{b}+\frac{1}{c} \right)=\frac{1}{6}.16=\frac{8}{3} \\ \end{align}\)

Vậy \(\frac{1}{3a+2b+c}+\frac{1}{a+3b+2c}+\frac{1}{2a+b+3c}\le \frac{8}{3}\).

Dấu bằng xảy ra \(\Leftrightarrow \left\{ \begin{align}  & a+b=b+c=c+a \\  & a=b=c \\  & \frac{1}{a}+\frac{1}{b}+\frac{1}{c}=16 \\ \end{align} \right.\Leftrightarrow a=b=c=\frac{3}{16}\)