Question

Cho a,b>0 thỏa mãn a+b=1 . Tìm GTNN của A =\(\left(a+\frac{1}{a}\right)^2+\left(b+\frac{1}{b}\right)^2\)

Answer 1

Do a ; b > 0 , áp dụng BĐT Cô - si cho 2 số dương , ta có :

\(A=\left(a+\frac{1}{a}\right)^2+\left(b+\frac{1}{b}\right)^2\ge2\left(a+\frac{1}{a}\right)\left(b+\frac{1}{b}\right)\)

\(\Rightarrow2\left[\left(a+\frac{1}{a}\right)^2+\left(b+\frac{1}{b}\right)^2\right]\ge\left(a+\frac{1}{a}+b+\frac{1}{b}\right)^2\)

\(\Rightarrow2A\ge\left(1+\frac{1}{a}+\frac{1}{b}\right)^2\)

Vì a ; b > 0 \(\Rightarrow\frac{1}{a}+\frac{1}{b}\ge\frac{4}{a+b}\)

\(\Rightarrow2A\ge\left(1+\frac{4}{a+b}\right)^2=\left(1+4\right)^2=25\)

\(\Rightarrow A\ge\frac{25}{2}\)

Dấu " = " xảy ra \(\Leftrightarrow a=b=\frac{1}{2}\)