Question

c/m voi 3 so thuc tuy y ta co

a² cong b² cong 1 lon hon hoac bang ab cong a cong b

Answer 1

Ta có: a²+b²+1≥ab+a+b

<=>2a²+2b²+2≥2ab+2a+2b

<=>(a²−2ab+b²)+(a²−2a+1)+(b²−2b+1)≥0

<=>(a−b)²+(a−1)²+(b−1)²≥0 ( Luôn đúng với V a,b)

Vậy  a²+b²+1≥ab+a+b