Question

CMR : \(a^2+b^2\ge\dfrac{1}{2}\)biết a+b = 1

Answer 1

Do: a+b=1

=> a=1-b

Thay a=1-b vào bất đẳng thức ta có :

(1-b)² + b² \(\ge\) \(\dfrac{1}{2}\)

\(\Rightarrow\) 1-2b+b² + b²\(\ge\)\(\dfrac{1}{2}\)

\(\Rightarrow\) 2- 4b+4b² \(\ge\) 1

Answer 2

a²+b²\(\ge\)2ab(theo cosi)

<=>2(a²+b²)\(\ge\)a²+b²+2ab

<=>2(a²+b²)\(\ge\)(a+b)²=1²=1(do a+b=1)

<=>a²+b²\(\ge\)\(\dfrac{1}{2}\)(đpcm)

Answer 3

Ace Legona,Hung nguyen,Hoang Hung Quan,Xuân Tuấn Trịnh......

Answer 4

ta có
(a-b)²≥0

⇔ a²-2ab+b²≥0

⇔ a²+b²≥2ab

⇔ 2a²+2b²≥a²+2ab+b²

⇔ 2(a²+b²)≥(a+b)²

⇔ 2(a²+b²)≥ 1

⇔ a²+b²≥0.5