Question

[Phạm Tuấn Kiệt]

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

[trần thành đạt]

Áp dụng BĐT Cauchy-Schwarz ta có

BT\(\ge\)\(\frac{\left(1+1+1\right)^2}{ab+bc+ac}+\frac{1}{a^2+b^2+c^2}=\frac{9}{ab+bc+ac}+\frac{1}{a^2+b^2+c^2}\)

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

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

Ta lại có ab+bc+ac =< (a+b+c)^2/3 =3

\(\Rightarrow BT\ge1+\frac{7}{3}=\frac{10}{3}\)

Vậy GTNN là \(\frac{10}{3}\)khi a=b=c=1

Answer 2

[pham trung thanh]

Cô-si Schwarzt dạng Engel là đc