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

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

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

Also we apply the square mean inequality to have

Since a+b = 1, we have

Combining all of them we have

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