Question

[Lê Đức Mạnh]

Cho a² >_ 2. Tifm GTNN cua bt:

A= a²+1/a²

Answer 1

[Hung nguyen]

\(A=a^2+\dfrac{1}{a^2}=\dfrac{3a^2}{4}+\left(\dfrac{a^2}{4}+\dfrac{1}{a^2}\right)\ge\dfrac{3.2}{4}+1=\dfrac{5}{2}\)

Vậy GTNN là \(A=\dfrac{5}{2}\) dấu = xảy ra khi \(a^2=2\)

Answer 2

[Ha Hoang Vu Nhat]

Ta có: \(A=a^2+\dfrac{1}{a^2}=\dfrac{3a^2}{4}+\dfrac{a^2}{4}+\dfrac{1}{a^2}=\dfrac{3a^2}{4}+\left(\dfrac{a^2}{4}+\dfrac{1}{a^2}\right)\)

Do \(a^2\ge2\) => \(\dfrac{3a^2}{4}\ge\dfrac{3}{4}.2=\dfrac{3}{2}\) (*)

Áp dụng BĐT cô-si :

\(\dfrac{a^2}{4}+\dfrac{1}{a^2}\ge2\sqrt{\dfrac{a^2}{4}.\dfrac{1}{a^2}}=2.\dfrac{1}{2}=1\) (**)

Từ (*) và (**) suy ra :

\(\dfrac{3a^2}{4}+\left(\dfrac{a^2}{4}+\dfrac{1}{a^2}\right)\ge\dfrac{3}{2}+1=\dfrac{5}{2}\)

<=> \(A\ge\dfrac{5}{2}\)

Dấu "=" xảy ra khi \(a^2=2\) <=> \(a=\pm\sqrt{2}\)

Vậy GTNN của \(A=a^2+\dfrac{1}{a^2}\) là \(\dfrac{5}{2}\) khi \(a=\pm\sqrt{2}\)