Xét
\(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\ge\frac{9}{a+b+c}\) \(\left(II\right)\)
\(\Leftrightarrow\left(a+b+c\right)\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)\ge9\)
\(\Leftrightarrow\frac{a}{a}+\frac{a}{b}+\frac{a}{c}+\frac{b}{a}+\frac{b}{b}+\frac{b}{c}+\frac{c}{a}+\frac{c}{b}+\frac{c}{c}\ge9\)
\(\Leftrightarrow1+1+1+\frac{a}{b}+\frac{a}{c}+\frac{b}{a}+\frac{b}{c}++\frac{c}{a}+\frac{c}{b}\ge9\)
\(\Leftrightarrow\left(\frac{a}{b}+\frac{b}{a}\right)+\left(\frac{a}{c}+\frac{c}{a}\right)+\left(\frac{b}{c}+\frac{c}{b}\right)\ge6\)
Áp dụng BĐT Cô - si cho các số dương ta có :
+) \(\frac{a}{b}+\frac{b}{a}\ge2\sqrt{\frac{a}{b}.\frac{b}{a}}=2\left(1\right)\)
+) \(\frac{a}{c}+\frac{c}{a}\ge2\sqrt{\frac{a}{c}.\frac{c}{a}}=2\left(2\right)\)
+) \(\frac{b}{c}+\frac{c}{b}\ge2\sqrt{\frac{b}{c}.\frac{c}{b}}=2\left(3\right)\)
Cộng vế với vế của các BĐT \(\left(1\right),\left(2\right),\left(3\right)\) ta có :
\(\left(\frac{a}{b}+\frac{b}{a}\right)+\left(\frac{a}{c}+\frac{c}{a}\right)+\left(\frac{b}{c}+\frac{c}{b}\right)\ge2+2+2\)
\(\Leftrightarrow\left(\frac{a}{b}+\frac{b}{a}\right)+\left(\frac{a}{c}+\frac{c}{a}\right)+\left(\frac{c}{b}+\frac{b}{c}\right)\ge6\left(I\right)\)
Vì các BĐT \(\left(I\right);\left(II\right)\) là tương đương nên BĐT (I) luôn đúng \(\Leftrightarrow\) BĐT (2) luôn đúng
Dấu "=" xảy ra \(\) \(\Leftrightarrow a=b=c\)
Vậy \(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\ge\frac{9}{a+b+c}\left(đpcm\right)\)
Cái này điều kiện phải là a,b,c dương
\(\Leftrightarrow\left(a+b+c\right)\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)\ge9\)
\(VT=3+\frac{a}{b}+\frac{a}{c}+\frac{b}{a}+\frac{b}{c}+\frac{c}{a}+\frac{c}{b}\ge3+2\sqrt{\frac{a}{b}.\frac{b}{a}}+2\sqrt{\frac{b}{c}.\frac{c}{b}}+2\sqrt{\frac{c}{a}.\frac{a}{c}}\)
\(=3+2+2+2=9\left(đpcm\right)\)
\("="\Leftrightarrow a=b=c\)
