\(S=\dfrac{a+b}{c}+\dfrac{b+c}{a}+\dfrac{c+a}{b}\)
\(S=\left(\dfrac{a}{c}+\dfrac{c}{a}\right)+\left(\dfrac{b}{c}+\dfrac{c}{b}\right)+\left(\dfrac{b}{a}+\dfrac{a}{b}\right)\)
Áp dụng bất đẳng thức Cauchy - Schwarz
\(\Rightarrow\left\{{}\begin{matrix}\dfrac{a}{c}+\dfrac{c}{a}\ge2\sqrt{\dfrac{ac}{ca}}=2\\\dfrac{b}{c}+\dfrac{c}{b}\ge2\sqrt{\dfrac{bc}{cb}}=2\\\dfrac{b}{a}+\dfrac{a}{b}\ge2\sqrt{\dfrac{ab}{ba}}=2\end{matrix}\right.\)
\(\Rightarrow\left(\dfrac{a}{c}+\dfrac{c}{a}\right)+\left(\dfrac{b}{c}+\dfrac{c}{b}\right)+\left(\dfrac{b}{a}+\dfrac{a}{b}\right)\ge2+2+2=6\)
\(\Leftrightarrow\dfrac{a+b}{c}+\dfrac{b+c}{a}+\dfrac{c+a}{b}\ge6\)
\(\Leftrightarrow S\ge6\) ( đpcm )
\(\Rightarrow S_{min}=6\)
Dấu " = " xảy ra khi \(a=b=c\)
cách 1 sử dụng BĐT
a)
\(S=\dfrac{a+b}{c}+\dfrac{b+c}{a}+\dfrac{c+a}{b}=\left(\dfrac{a}{c}+\dfrac{b}{c}+\dfrac{b}{a}+\dfrac{c}{a}+\dfrac{c}{b}+\dfrac{a}{b}\right)\)đã áp cô_si --> áp tới bến luôn
\(S=\left(\dfrac{a}{c}+\dfrac{b}{c}+\dfrac{b}{a}+\dfrac{c}{a}+\dfrac{c}{b}+\dfrac{a}{b}\right)\ge6\sqrt[6]{\dfrac{\left(abc\right)^2}{\left(abc\right)^2}}=6\) =>dpcm
b) min S=6
khi \(\dfrac{a}{b}=\dfrac{b}{a}=\dfrac{c}{a}=\dfrac{a}{c}=\dfrac{b}{c}=\dfrac{c}{b}\Rightarrow a=b=c\)
cách2sử dụng HĐT \(\left(x-y\right)^2\ge0\forall x,y\)
\(S=\left(\dfrac{a}{b}-2+\dfrac{b}{a}\right)+\left(\dfrac{c}{b}-2+\dfrac{b}{c}\right)+\left(\dfrac{a}{c}-2+\dfrac{c}{a}\right)+6\)
\(S=\left(\sqrt{\dfrac{c}{b}}-\sqrt{\dfrac{b}{c}}\right)^2+\left(\sqrt{\dfrac{a}{b}}-\sqrt{\dfrac{b}{a}}\right)^2+\left(\sqrt{\dfrac{a}{c}}-\sqrt{\dfrac{c}{a}}\right)^2+6\ge6\)=> dpcm
Min S=6
khi \(\left\{{}\begin{matrix}\left(\sqrt{\dfrac{c}{b}}-\sqrt{\dfrac{b}{c}}\right)=0\\\left(\sqrt{\dfrac{c}{b}}-\sqrt{\dfrac{b}{c}}\right)=0\\\left(\sqrt{\dfrac{c}{b}}-\sqrt{\dfrac{b}{c}}\right)=0\end{matrix}\right.\)\(\Rightarrow a=b=c\)
