Problem

Prove that if \(p, q \gt 0\) and \(p + q = 1\), then

\[ (p + \frac{1}{p})^2 + (q + \frac{1}{q})^2 \geq 25/2\]

Proof

Since \(p, q \gt 0\) therefore \(p + \frac{1}{p} \gt 0 \) and \(q + \frac{1}{q} \gt 0 \). As \(QM \geq AM\) for positive numbers:

\[ \sqrt{\frac{(p + \frac{1}{p})^2 + (q + \frac{1}{q})^2}{2}} \geq \frac{(p + \frac{1}{p}) + (q + \frac{1}{q})}{2}\]

Squaring both sides and rearranging:

\[\implies \frac{(p + \frac{1}{p})^2 + (q + \frac{1}{q})^2}{2} \geq [\frac{(p + q) + (\frac{1}{p} + \frac{1}{q})}{2}]^2\] \[\implies (p + \frac{1}{p})^2 + (q + \frac{1}{q})^2 \geq \frac{1}{2}[(p + q) + (\frac{1}{p} + \frac{1}{q})]^2\] \[\implies (p + \frac{1}{q})^2 + (q + \frac{1}{q})^2 \geq \frac{1}{2}[1 + (\frac{1}{p} + \frac{1}{q})]^2 \tag{1} \]

Since \(AM \geq HM\), we have:

\[ \frac{p+q}{2} \geq \frac{2}{\frac{1}{p} + \frac{1}{q}} \]

Using the fact that \(p + q = 1\) and rearranging:

\[ \frac{1}{p} + \frac{1}{q} \geq 4 \] \[ \implies 1+ (\frac{1}{p} + \frac{1}{q}) \geq 5 \] \[ \implies [1+ (\frac{1}{p} + \frac{1}{q})]^2 \geq 25 \] \[ \implies \frac{1}{2}[1+ (\frac{1}{p} + \frac{1}{q})]^2 \geq \frac{25}{2} \tag{2} \]

From inequalities \((1)\) and \((2):\)

\[ (p + \frac{1}{p})^2 + (q + \frac{1}{q})^2 \geq 25/2 \]

Hence, proved.

Reference

Problem 1.15 of Inequalities by Ozgur Kircak