Cho các số thực dương a, b, c thỏa mãn \(c\ge a\)....

- Hướng dẫn giải

Phương pháp giải:

Giải chi tiết:

Áp dụng BĐT Cauchy ta có:

\(\begin{align}  & {{\left( \frac{a}{a+b} \right)}^{2}}+\frac{1}{4}\ge 2.\frac{1}{2}.\frac{a}{a+b}=\frac{a}{a+b} \\  & {{\left( \frac{b}{b+c} \right)}^{2}}+\frac{1}{4}\ge 2.\frac{1}{2}.\frac{b}{b+c}=\frac{b}{b+c} \\  & {{\left( \frac{c}{c+a} \right)}^{2}}+\frac{1}{4}\ge 2.\frac{1}{2}.\frac{c}{c+a}=\frac{c}{c+a} \\  & \Rightarrow {{\left( \frac{a}{a+b} \right)}^{2}}+{{\left( \frac{b}{b+c} \right)}^{2}}+4{{\left( \frac{c}{c+a} \right)}^{2}}+\frac{3}{2}\ge \frac{a}{a+b}+\frac{b}{b+c}+\frac{4c}{c+a} \\  & \Leftrightarrow {{\left( \frac{a}{a+b} \right)}^{2}}+{{\left( \frac{b}{b+c} \right)}^{2}}+4{{\left( \frac{c}{c+a} \right)}^{2}}\ge \frac{a}{a+b}+\frac{b}{b+c}+\frac{4c}{c+a}-\frac{3}{2} \\ \end{align}\)

Ta chứng minh bổ đề \(\frac{1}{x+1}+\frac{1}{y+1}\ge \frac{2}{\sqrt{xy}+1}\,\,\forall xy\ge 1\)

\(\begin{align}  & \frac{1}{x+1}+\frac{1}{y+1}\ge \frac{2}{\sqrt{xy}+1} \\  & \Leftrightarrow \frac{x+y+2}{xy+x+y+1}\ge \frac{2}{\sqrt{xy}+1} \\  & \Leftrightarrow \frac{x+y+2}{xy+x+y+1}-\frac{2}{\sqrt{xy}+1}\ge 0 \\  & \Leftrightarrow \left( x+y+2 \right)\left( \sqrt{xy}+1 \right)-2\left( xy+x+y+1 \right)\ge 0 \\  & \Leftrightarrow \left( x+y+2 \right)\sqrt{xy}+x+y+2-2xy-2x-2y-2\ge 0 \\  & \Leftrightarrow \left( x+y+2 \right)\sqrt{xy}-x-y-2xy\ge 0 \\  & \Leftrightarrow \left( x+y+2 \right)\sqrt{xy}-\left( x+y+2 \right)+2-2xy\ge 0 \\  & \Leftrightarrow \left( x+y+2 \right)\left( \sqrt{xy}-1 \right)-2\left( xy-1 \right)\ge 0 \\  & \Leftrightarrow \left( \sqrt{xy}-1 \right)\left( x+y+2-2\left( \sqrt{xy}+1 \right) \right)\ge 0 \\  & \Leftrightarrow \left( \sqrt{xy}-1 \right)\left( x+y-2\sqrt{xy} \right)\ge 0 \\  & \Leftrightarrow \left( \sqrt{xy}-1 \right){{\left( \sqrt{x}-\sqrt{y} \right)}^{2}}\ge 0\,\,\left( luon\,\,dung\,\forall xy\ge 1 \right) \\ \end{align}\)

Áp dụng bổ để trên ta có : \(\frac{a}{a+b}+\frac{b}{b+c}=\frac{1}{1+\frac{b}{a}}+\frac{1}{1+\frac{c}{a}}\ge \frac{2}{\sqrt{\frac{c}{a}}+1}\)

\(\Rightarrow \frac{a}{a+b}+\frac{b}{b+c}+\frac{4c}{c+a}-\frac{3}{2}\ge \frac{2}{\sqrt{\frac{c}{a}}+1}+\frac{4c}{c+a}-\frac{3}{2}\)

Ta cần chứng minh \(\frac{2}{\sqrt{\frac{c}{a}}+1}+\frac{4c}{c+a}-\frac{3}{2}\ge \frac{3}{2}\Leftrightarrow \frac{2}{\sqrt{\frac{c}{a}}+1}+\frac{4}{1+\frac{a}{c}}\ge 3\,\,\left( * \right)\)

Đặt \(t=\sqrt{\frac{a}{c}}\Rightarrow 0<t\le 1\,\,\left( Do\,\,c\ge a \right)\)

 Khi đó :

\(\begin{align}  & \left( * \right)\Leftrightarrow \frac{2}{\frac{1}{t}+1}+\frac{4}{1+{{t}^{2}}}\ge 3 \\  & \Leftrightarrow \frac{2t}{t+1}+\frac{4}{1+{{t}^{2}}}\ge 3 \\  & \Leftrightarrow 2t\left( 1+{{t}^{3}} \right)+4\left( 1+t \right)-3\left( 1+t \right)\left( 1+{{t}^{2}} \right)\ge 0 \\  & \Leftrightarrow 2t+2{{t}^{3}}+4+4t-3-3{{t}^{2}}-3t-3{{t}^{3}}\ge 0 \\  & \Leftrightarrow -{{t}^{3}}-3{{t}^{2}}+3t+1\ge 0 \\  & \Leftrightarrow \left( 1-t \right)\left( 1+t+{{t}^{2}} \right)+3t\left( 1-t \right)\ge 0 \\  & \Leftrightarrow \left( 1-t \right)\left( {{t}^{2}}+4t+1 \right)\ge 0 \\ \end{align}\)

Ta có \(0<t\le 1\Rightarrow \left\{ \begin{align}  & {{t}^{2}}+4t+1>0 \\  & 1-t\ge 0 \\ \end{align} \right.\Rightarrow \left( 1-t \right)\left( {{t}^{2}}+4t+1 \right)\ge 0\)

Do đó \(\frac{2}{\sqrt{\frac{c}{a}}+1}+\frac{4c}{c+a}-\frac{3}{2}\ge \frac{3}{2}\) .

Vậy \({{\left( \frac{a}{a+b} \right)}^{2}}+{{\left( \frac{b}{b+c} \right)}^{2}}+4{{\left( \frac{c}{c+a} \right)}^{2}}\ge \frac{3}{2}\).