Question

[༺ ๖ۣۜPhạm ✌Tuấn ✌Kiệτ ༻]

Cho a, b, c > 0 thoa man a + b + c = 3.

Tim GTNN : \(\dfrac{1}{ab}+\dfrac{1}{bc}+\dfrac{1}{ac}+\dfrac{1}{a^2+b^2+c^2}\)

Answer 1

[Nguyễn Hải Đăng]

Áp dụng bất đẳng thức Cauchy-Schwarz:

\(VT=\dfrac{1}{ab}+\dfrac{1}{bc}+\dfrac{1}{ac}+\dfrac{1}{a^2+b^2+c^2}\ge\dfrac{\left(1+1+1\right)^2}{ab+bc+ac}+\dfrac{1}{a^2+b^2+c^2}\)

\(=\dfrac{9}{ab+bc+ac}+\dfrac{1}{a^2+b^2+c^2}\)

\(=\dfrac{1}{ab+bc+ac}+\dfrac{1}{ab+bc+ac}+\dfrac{7}{ab+bc+ac}+\dfrac{1}{a^2+b^2+c^2}\)

\(\ge\dfrac{\left(1+1+1\right)^2}{ab+bc+ac+ab+bc+ac+a^2+b^2+c^2}+\dfrac{7}{ab+bc+ac}\)

\(=\dfrac{9}{\left(a+b+c\right)^2}+\dfrac{7}{ab+bc+ac}\)

Áp dụng bất đẳng thức AM-GM cho 2 số dương:

\(ab+bc+ac\le\dfrac{\left(a+b+c\right)^2}{3}=\dfrac{1^2}{3}=\dfrac{1}{3}\)

Ta có: \(\dfrac{9}{\left(a+b+c\right)^2}+\dfrac{7}{ab+bc+ac}\ge\dfrac{9}{\left(a+b+c\right)^2}+\dfrac{7}{\dfrac{1}{3}}=9+21=30\)