Question

Cho 2 số a,b \(\ne0\) thỏa mãn

\(\left(ab+6\right)^2=\left(a^2+4\right)\left(b^2+9\right)\). Tính \(A=\dfrac{a^2-ab+b^2}{a^2+ab+b^2}\)

Answer 1

Áp dụng BĐT Cauchy-Schwarz ta có:

\(VT=\left(a^2+4\right)\left(b^2+9\right)\)

\(\ge\left(\sqrt{a^2b^2}+\sqrt{4\cdot9}\right)^2=\left(ab+36\right)^2=VP\)

Xảy ra khi \(\dfrac{a^2}{4}=\dfrac{b^2}{9}\Rightarrow\dfrac{a}{2}=\dfrac{b}{3}\Rightarrow b=\dfrac{3a}{2}\)

Khi đó \(A=\dfrac{a^2-ab+b^2}{a^2+ab+b^2}=\dfrac{a^2-a\cdot\dfrac{3a}{2}+\left(\dfrac{3a}{2}\right)^2}{a^2+a\cdot\dfrac{3a}{2}+\left(\dfrac{3a}{2}\right)^2}=\dfrac{7}{19}\)