Question

Bài 5.

a, Chứng minh (x + y + z)² ≥ 3(xy + yz + xz)

b, Với a,b,c > 0 thỏa mãn a + b + c = 3. Tìm giá trị nhỏ nhất của Q = \(\frac{a}{b^2+1}+\frac{b}{c^2+1}+\frac{c}{a^2+1}\)

Answer 1

a) Ta có: \(\left(x+y+z\right)^2\ge3\left(xy+yz+xz\right)\)

\(\Leftrightarrow x^2+y^2+z^2\ge xy+yz+xz\)

\(\Leftrightarrow\left(x-y\right)^2+\left(y-z\right)^2+\left(z-x\right)^2\ge0\) (luôn đúng)

\("="\Leftrightarrow x=y=z\)

b)

\(Q=\frac{a}{b^2+1}+\frac{b}{c^2+1}+\frac{c}{a^2+1}=a-\frac{ab^2}{b^2+1}+b-\frac{bc^2}{c^2+1}+c-\frac{ca^2}{a^2+1}\ge a+b+c-\frac{a}{2}-\frac{b}{2}-\frac{c}{2}=\frac{3}{2}\)

\("="\Leftrightarrow a=b=c=1\)

Answer 2

b) \(Q=a-\frac{ab^2}{b^2+1}+b-\frac{bc^2}{c^2+1}+c-\frac{ca^2}{a^2+1}\)

\(\ge\left(a+b+c\right)-\left(\frac{ab^2}{2b}+\frac{bc^2}{2c}+\frac{ca^2}{2a}\right)\)

\(=\left(a+b+c\right)-\left(\frac{ab}{2}+\frac{bc}{2}+\frac{ca}{2}\right)\)

\(\ge a+b+c-\frac{\frac{\left(a+b+c\right)^2}{3}}{2}=3-\frac{3}{2}=\frac{3}{2}\)

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