Question

Chứng​ minh rằng​ nếu: a + b =1 thì​ a^​2 ​+ b^​2​ ≥ \(\dfrac{1}{2}\)

Answer 1

Ta có:

\(\left(a-b\right)^2\ge0\)

\(\Leftrightarrow a^2-2ab+b^2\ge0\)

\(\Leftrightarrow a^2+b^2\ge2ab\)

\(\Leftrightarrow a^2+b^2+a^2+b^2\ge a^2+b^2+2ab\)

\(\Leftrightarrow2\left(a^2+b^2\right)\ge\left(a+b\right)^2\)

\(\Leftrightarrow2\left(a^2+b^2\right)\ge1\)

\(\Leftrightarrow a^2+b^2\ge\dfrac{1}{2}\)

=> ĐPCM

Answer 2

Ta có: a + b = 1

<=> \(\left(a+b\right)^2=1\)

<=> \(a^2+2ab+b^2=1\) (1)

Lại có: \(\left(a-b\right)^2\ge0\)

<=> \(a^2-2ab+b^2\ge0\) (2)

Cộng (1) và (2) vế theo vế ta được:

\(2a^2+2b^2\ge1\)

<=>\(2\left(a^2+b^2\right)\ge1\)

\(\Leftrightarrow a^2+b^2\ge\dfrac{1}{2}\) (đpcm)