*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 \(\displaystyle {\left( a + \frac{1}{a} \right)^2 + \left(b + \frac{1}{b} \right)^2 \ge 12 \frac{1}{2} = \frac{25}{2}} \)

*Related*

Other way to solve the problem is: The given expression equal to (a-b)^2 + (1/a-1/b)^2 +2ab+2/ab +4. Least value is obtained if both (a-b)^2 & (1/a-1/b)^2 equal 0, or a=b. Also a+b=1. Hence a=b=1/2. Hence we get the given expression greater or equal to 2*1/2*1/2 +2*4 +4= 25/2

WE may solve this questino using cauchy-schwarz inequality.

[(a+b){(1/a)+(1/b)}] >={(a*1/a)+(b*1/b)}^2

or,{(1/a)+(1/b)}>=4 ,[as a+b=1]

now we use the rms-am inequality,

[{(a+1/a)^2}+{(b+1/b)^2}]>=1/2[{a+1/a+b+1/b}^2]

or,[{(a+1/a)^2}+{(b+1/b)^2}]>=1/2{(1+4)^2}

or,[{(a+1/a)^2}+{(b+1/b)^2}]>=25/2