\(A=a^2+b^2+\frac{1}{a^2}+\frac{1}{b^2}\)
\(=a^2+b^2+\frac{b^2+a^2}{a^2b^2}\ge0\)
\(MinA=0\Leftrightarrow\hept{\begin{cases}a^2=0\\b^2=0\end{cases}\Rightarrow\hept{\begin{cases}a=0\\b=0\end{cases}}}\)
Từ giả thuyết ta có \(\hept{\begin{cases}\frac{1}{4}\ge ab\\ab+\frac{1}{16ab}\ge\frac{1}{2}\end{cases}}\)
Ta có \(\frac{1}{a^2}+\frac{1}{b^2}+a^2+b^2\)\(\ge\)\(2\left(ab+\frac{1}{ab}\right)\)
= \(2\left(ab+\frac{1}{16ab}+\frac{15}{16ab}\right)\)
\(\ge\)\(2\left(\frac{1}{2}+\frac{15}{16×\frac{1}{4}}\right)\)= \(\frac{17}{2}\)
Vậy GTNN là \(\frac{17}{2}\)đạt được khi a = b = \(\frac{1}{2}\)
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