Question

[pham thuy trang]

chứng minh với mọi số a,b  ta có a² + 5b² -4ab + 2a - 6b + 3 > 0

Answer 1

[Hoàng Phúc]

\(a^2+5b^2-4ab+2a-6b+3\)

\(=a^2-4ab+2a+5b^2-6b+3\)

\(=a^2-2a\left(2b-1\right)+5b^2-6b+3\)

\(=a^2-2.a.\frac{2b-1}{2}+\left(\frac{2b-1}{2}\right)^2+5b^2-6b-\left(\frac{2b-1}{2}\right)^2+3\)

\(=\left(a-\frac{2b-1}{2}\right)^2+5a^2-6b-\frac{\left(2b-1\right)^2}{4}+3\)

\(=\left(a-\frac{2b-1}{2}\right)^2+5a^2-6b-\frac{4b^2-4b+1}{4}+3\)

\(=\left(a-\frac{2b-1}{2}\right)^2+5a^2-6b-b^2+b-\frac{1}{4}+3\)

\(=\left(a-\frac{2b-1}{2}\right)^2+4b^2-5b+\frac{11}{4}\)

\(=\left(a-\frac{2b-1}{2}\right)^2+\left(2b\right)^2-2.2b.\frac{5}{4}+\frac{25}{16}+\frac{19}{16}\)

\(=\left(a-\frac{2b-1}{2}\right)^2+\left(2b-\frac{5}{4}\right)^2+\frac{19}{16}\)

Vì \(\left(a-\frac{2b-1}{2}\right)^2\ge0;\left(2b-\frac{5}{4}\right)^2\ge0=>\left(a-\frac{2b-1}{2}\right)^2+\left(2b-\frac{5}{4}\right)^2+\frac{19}{16}\ge\frac{19}{16}>0\) (với mọi a,b)  (đpcm)