Question

Cho a,b,c >0 thỏa mãn a+b+c\(\le\)\(\frac{3}{2}\).Chứng minh

a,\(\frac{1}{a}\)+\(\frac{1}{b}\)+\(\frac{1}{c}\)\(\ge\)6

b,a+ b+ c+ \(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\)\(\ge\)\(\frac{15}{2}\)

Answer 1

\(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\ge6\)

=> \(-\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)\le-6\)

=> \(-\left(a+b+c\right)\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)\le-6.\frac{3}{2}\)

=> \(\left(a+b+c\right)\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)\ge9\)

=> \(1+\frac{a}{b}+\frac{a}{c}+\frac{b}{a}+1+\frac{b}{c}+\frac{c}{a}+\frac{c}{b}+1\ge9\)

=> \(\left(\frac{a}{b}+\frac{b}{a}\right)+\left(\frac{a}{c}+\frac{c}{a}\right)+\left(\frac{b}{c}+\frac{c}{b}\right)\ge6\)(1)

Dễ thấy \(\frac{a}{b}+\frac{b}{a}\ge2\)(với a,b > 0)

=> (1) đúng 

=> BĐTđược chứng minh

Answer 2

b)Đặt  \(A=a+b+c+\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\left(a,b,c>0\right)\).

\(A=4\left(a+b+c\right)-3\left(a+b+c\right)+\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\).

\(A=\left(4a+\frac{1}{a}\right)+\left(4b+\frac{1}{b}\right)+\left(4c+\frac{1}{c}\right)-3\left(a+b+c\right)\).

Vì \(a>0\)nên áp dụng bất đẳng thức Cô-si cho 2 số dương, ta được:

\(4a+\frac{1}{a}\ge2\sqrt{4.a.\frac{1}{a}}=4\left(1\right)\).

Dấu bằng xảy ra \(\Leftrightarrow4a=\frac{1}{a}\Leftrightarrow a=\frac{1}{2}\).

 Chứng minh tương tự, ta được:

\(4b+\frac{1}{b}\ge4\left(b>0\right)\left(2\right)\).
Dấu bằng xảy ra \(\Leftrightarrow b=\frac{1}{2}\).

Chứng minh tương tự, ta được:

\(4c+\frac{1}{c}\ge4\left(c>0\right)\left(3\right)\).
Dấu bằng xảy ra \(\Leftrightarrow c=\frac{1}{2}\).

Từ \(\left(1\right),\left(2\right),\left(3\right)\), ta được:

\(\left(4a+\frac{1}{a}\right)+\left(4b+\frac{1}{b}\right)+\left(4c+\frac{1}{c}\right)\ge4+4+4=12\).

\(\Leftrightarrow\left(4a+\frac{1}{a}\right)+\left(4b+\frac{1}{b}\right)+\left(4c+\frac{1}{c}\right)-3\left(a+b+c\right)\ge\)\(12-3\left(a+b+c\right)\).

\(\Leftrightarrow A\ge12-3\left(a+b+c\right)\left(4\right)\).

Mặt khác, ta có: \(a+b+c\le\frac{3}{2}\).

\(\Leftrightarrow3\left(a+b+c\right)\le\frac{9}{2}\).

\(\Rightarrow-3\left(a+b+c\right)\ge-\frac{9}{2}\).

\(\Leftrightarrow12-3\left(a+b+c\right)\ge\frac{15}{2}\left(5\right)\).
Dấu bằng xảy ra \(\Leftrightarrow a+b+c=\frac{3}{2}\).

Từ \(\left(4\right)\)và \(\left(5\right)\), ta được:

\(A\ge\frac{15}{2}\).

Dấu bằng xảy ra \(\Leftrightarrow a=b=c=\frac{1}{2}\).

Vậy với \(a,b,c>0\)và \(a+b+c\le\frac{3}{2}\)thì \(a+b+c+\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\ge\frac{15}{2}\).