Question

[Linh Luna]

Cho a/b > 0 chứng minh rằng a/b + b/a lớn hơn hoặc bằng 2

Answer 1

[Mới vô]

Gọi b = a + k (k \(\in\) Z, k \(\ne\) -a)

\(\dfrac{a}{b}>0\)

Ta có:

\(\dfrac{a}{a+k}+\dfrac{a+k}{a}\\ =\dfrac{a^2}{a\cdot\left(a+k\right)}+\dfrac{\left(a+k\right)^2}{a\cdot\left(a+k\right)}\\ =\dfrac{a^2+\left(a+k\right)^2}{a\cdot\left(a+k\right)}\\ =\dfrac{a^2+\left(a^2+2ak+k^2\right)}{a^2+ak}\\ =\dfrac{a^2+a^2+2ak+k^2}{a^2+ak}\\ =\dfrac{2a^2+2ak+k^2}{a^2+ak}\\ =\dfrac{2a^2+2ak}{a^2+ak}+\dfrac{k^2}{a^2+ak}\\ =\dfrac{2\cdot\left(a^2+ak\right)}{a^2+ak}+\dfrac{k^2}{a^2+ak}\\ =2+\dfrac{k^2}{a^2+ak}>2\)

Vậy \(\dfrac{a}{a+k}+\dfrac{a+k}{a}>2\Rightarrow\dfrac{a}{b}+\dfrac{b}{a}>2\left(đpcm\right)\)

Answer 2

[Hoang Hung Quan]

Sai! CMR: \(\dfrac{a}{b}+\dfrac{b}{a}\ge2\) mà?

Vào đây đi:

dfrac{a}{b}+\dfrac{b}{a - Hoc24