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If a and b are positive real numbers such that a + b = 1, prove that $\displaystyle \left( a + \frac{1}{a} \right)^2$ + $$\left( b + \frac{1}{b} \right)^2 \ge \frac{25}{2}$$

Discussion

Some theoretical background.

We replace 1 by a+b and expand the squares.

$\displaystyle \left( a + \frac{a+b}{a} \right)^2 + \left(b + \frac{a+b}{b} \right)^2$
$\displaystyle \left( a + \frac{b}{a} + 1 \right)^2 + \left(b + \frac{a}{b} + 1 \right)^2$
$\displaystyle \left( a^2 + \frac{b^2}{a^2} + 1 + 2a + 2b + 2\frac{b}{a} \right) + \left(b^2 + \frac{b^2}{a^2} + 1 + 2b + 2a + 2 \frac{a}{b} \right)$

Now we use A.M. – G.M. inequality to have

$\displaystyle { \frac {\frac{b^2}{a^2} + \frac{a^2}{b^2}}{2}\ge \sqrt {\frac{b^2}{a^2} \times \frac{a^2}{b^2}} = 1 }$
$\displaystyle { \frac {\frac{b}{a} + \frac{a}{b}}{2}\ge \sqrt {\frac{b}{a} \times \frac{a}{b}} = 1 }$

Also we apply the square mean inequality to have

$\displaystyle {\frac{a^2 + b^2}{2} \ge \left(\frac{a+b}{2} \right)^2}$ Since a+b = 1, we have $a^2 + b^2 \ge \frac{1}{2}$

Combining all of them we have $\displaystyle {\left( a + \frac{1}{a} \right)^2 + \left(b + \frac{1}{b} \right)^2 \ge 12 \frac{1}{2} = \frac{25}{2}}$