Dat \(\hept{\begin{cases}A=\frac{b+c}{a}+\frac{c+a}{b}+\frac{a+b}{c}\\B=\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}\end{cases}}\)
Ta co:\(A=\left(\frac{a}{b}+\frac{b}{a}\right)+\left(\frac{b}{c}+\frac{c}{b}\right)+\left(\frac{c}{a}+\frac{a}{c}\right)\ge2+2+2=6\left(1\right)\)
\(B=\frac{a+b+c}{a+b}+\frac{a+b+c}{b+c}+\frac{a+b+c}{c+a}-3\)
\(=\left(a+b+c\right)\left(\frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{c+a}\right)-3\ge\left(a+b+c\right).\frac{9}{2\left(a+b+c\right)}-3=\frac{3}{2}\left(2\right)\)
Cong ve voi ve cua (1) va (2) ta duoc:
\(P=A+B\ge6+\frac{3}{2}=\frac{15}{2}\)
Dau '=' xay ra khi \(a=b=c\)
Chứng minh ĐBT:\(\frac{b}{a}+\frac{a}{b}\ge2\left(a,b\ne0\right)\)(Dấu "="\(\Leftrightarrow a=b=1\))
Ta có: \(\left(a-b\right)^2\ge0\)
\(\Leftrightarrow a^2-2ab+b^2\ge0\)
\(\Leftrightarrow a^2+b^2\ge2ab\)
\(\Leftrightarrow\frac{a^2+b^2}{ab}\ge2\)
\(\Leftrightarrow\frac{a}{b}+\frac{b}{a}\ge2\left(đpcm\right)\)
Vậy \(\frac{b+c}{a}+\frac{a}{b+c}\ge2\)
\(\frac{a+c}{b}+\frac{b}{c+a}\ge2\)
\(\frac{a+b}{c}+\frac{c}{b+a}\ge2\)
\(\Rightarrow P\ge6\)
Vậy \(P_{min}=6\Leftrightarrow\hept{\begin{cases}a=b+c\\b=a+c\\c=a+b\end{cases}}\)
\(\frac{b+c}{4a}+\frac{a}{b+c}\ge1\)
Tương tự với 2 cái kia, cộng lại ta đc: \(P\ge3+\frac{3}{4}\left(\frac{b+c}{a}+\frac{c+a}{b}+\frac{a+b}{c}\right)=\frac{3}{4}+\frac{3}{4}\left(a+b+c\right)\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)\ge\frac{15}{2}\)
Dấu "=" xảy ra khi \(a=b=c\)
Áp dụng bất đẳng thức Cauchy-schwarz ta có:\(\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}=\frac{a^2}{ab+ac}+\frac{b^2}{cb+ab}+\frac{c^2}{ac+cb}\ge\frac{\left(a+b+c\right)^2}{2\left(ab+bc+ca\right)}=\frac{3}{2}\)( 1 )
Áp dụng BĐT phụ \(\left(a+b+c\right)^2\ge3\left(ab+bc+ca\right)\) vào ( 1 )
Lại có:\(\frac{b+c}{a}+\frac{c+a}{b}+\frac{a+b}{c}=\frac{b}{a}+\frac{a}{b}+\frac{c}{b}+\frac{b}{c}+\frac{a}{c}+\frac{c}{a}\ge2+2+2=6\) ( Cô-si )
Khi đó:\(P\ge\frac{3}{2}+6=\frac{15}{2}\)
Dấu bằng xảy ra tại \(a=b=c\)
