If a > 0 and b > 0 and a + b = 1, then \((a + \frac{1}{a})^2 + (b + \frac{1}{b})^2\) is 

 Concept:

• Let x₁, x₂, ..., x_n ∈ R then,

1. The arithmetic mean of x₁, x2, ..., x_n is defined by  \(A=\frac{ x_1+x_2+ ...+x_n}{n}\) 
2. The geometric mean of x1, x2, ..., x_n is defined by \(\sqrt[n]{x_1.x_2.x_3.\dots.x_n}\) 
3. The harmonic mean of x1, x2, ..., xn is defined by \(H=\frac{n}{\frac{1}{x_1}+\frac{1}{x_2}+...+\frac{1}{x_n}}\)

• ​Arithmetic-Geometric mean inequality: \(\sqrt[n]{x_1.x_2.x_3.\dots.x_n}\)  ≤ \(\frac{ x_1+x_2+ ...+x_n}{n}\)

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Calculations:

Given, a > 0 and b > 0 and a + b = 1, 

⇒ \(\frac{a+b}{2}=\frac{1}{2}\) 

⇒ \(\sqrt[]{ab}\)  \(≤\frac{a+b}{2}=\frac{1}{2}\) 

⇒ \(ab ≤ \frac{1}{4}\)

Now, a² + b² = (a + b)² - 2ab

\(≥ 1^2+2(\frac{-1}{4})\)   (∵ a + b = 1 and \(ab ≤ \frac{1}{4}\)  or  \(-ab ≥ \frac{-1}{4}\) )

\(≥ 1-\frac{1}{2}\)

\(≥ \frac{1}{2}\)

⇒ \((a^2+b^2)≥ \frac{1}{2}\)

Now consider,

\((a + \frac{1}{a})^2 + (b + \frac{1}{b})^2\)

\(=a^2+\frac{1}{a^2}+2+b^2+\frac{1}{b^2}+2\)

\(=(a^2+b^2)+(\frac{1}{a^2}+\frac{1}{b^2})+4\)

\(=(a^2+b^2)+(\frac{a^2+b^2}{a^2b^2})+4\)

\(≥ \frac{1}{2}+\frac{1}{2}\times 16+4\)   | ∵ \(ab ≤ \frac{1}{4}\) on squaring this we get \(a^2b^2 ≤ \frac{1}{16}\) and so \(\frac{1}{a^2b^2} ≥ 16\)  

∴ \((a + \frac{1}{a})^2 + (b + \frac{1}{b})^2≥\frac{25}{2}\) 

Hence, the correct answer is option 3)