Question

tìm giá trị nhỏ nhất của F= a^2 + ab + b^2 - 3a - 3b + 3

Answer 1

\(F=a^2+ab+b^2-3a-3b+3\)

\(=\left(a^2-2a+1\right)+\left(b^2-2b+1\right)+\left(ab-a-b+1\right)\)

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

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

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

Ta thấy \(\left[\left(a-1\right)+\frac{1}{2}\left(b-1\right)\right]^2\ge0\) và \(\frac{3}{4}\left(b-1\right)^2\ge0\) với mọi a;b

Nên \(A=\left[\left(a-1\right)+\frac{1}{2}\left(b-1\right)\right]^2+\frac{3}{4}\left(b-1\right)^2\ge0\forall a;b\) có GTNN là 0

Dấu "=" xảy ra \(\Leftrightarrow a=b=1\)

Answer 2

\(4F=4a^2+4ab+4b^2-12a-12b+12\)

\(=\left(4a^2+b^2+4+4ab-12a-6b\right)+\left(3b^2-6b+3\right)\)

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

vì \(\left(2a+b-2\right)^2\ge0\forall a,b\)

\(3\left(b-1\right)^2\ge0\forall b\)

\(\Rightarrow4F\ge0\forall a,b\Rightarrow F\ge0\forall a,b\)

\(\Rightarrow GTNN\)của F là 0 \(\Leftrightarrow\hept{\begin{cases}b-1=0\\2a+b-3=0\end{cases}}\Rightarrow\hept{\begin{cases}b=1\\a=1\end{cases}}\)

Answer 3

F=a2+ab+b2−3a−3b+3

=(�2−2�+1)+(�2−2�+1)+(��−�−�+1)=(a2−2a+1)+(b2−2b+1)+(ab−a−b+1)

=(�−1)2+(�−1)2+(�−1)(�−1)=(a−1)2+(b−1)2+(a−1)(b−1)

=[(�−1)2+(�−1)(�−1)+14(�−1)2]+34(�−1)2=[(a−1)2+(a−1)(b−1)+41​(b−1)2]+43​(b−1)2

=[(�−1)+12(�−1)]2+34(�−1)2=[(a−1)+21​(b−1)]2+43​(b−1)2

Ta thấy [(�−1)+12(�−1)]2≥0[(a−1)+21​(b−1)]2≥0 và 34(�−1)2≥043​(b−1)2≥0 với mọi a;b

Nên �=[(�−1)+12(�−1)]2+34(�−1)2≥0∀�;�A=[(a−1)+21​(b−1)]2+43​(b−1)2≥0∀a;b có GTNN là 0

Dấu "=" xảy ra $\iff a=b=1\; =)$