Question

If a, b are positive real variables whose sum is a constant λ, then the minimum value of p(1 + 1/a)(1 + 1/b) is

Answer 1

Given a+b=Ca+b=C so b=C−ab=C−a
   f=(1+1a)(1+1b)−−−−−−−−−−−−√f=(1+1a)(1+1b)
   g=f2=(1+1a)(1+1C−a)g=f2=(1+1a)(1+1C−a)
Taking the derivative of g wrt a and equating to 0
   g^' = (-\frac{1}{a^2})(1+\frac{1}{C-a})g^' = (-\frac{1}{a^2})(1+\frac{1}{C-a})
        +(1+1a)(1(C−a)2)=0+(1+1a)(1(C−a)2)=0
Simplifying the above gives
   (C−2a)(C+1)=0(C−2a)(C+1)=0 or a=C2a=C2

Taking the second derivative of g wrt a
   g′′=(2a3)(1+1C−a)g″=(2a3)(1+1C−a)
        −1a21(C−a)2−1a21(C−a)2
        +(1+1a)(2(C−a)3)+(1+1a)(2(C−a)3)
        −1a2(1(C−a)2)−1a2(1(C−a)2)
at a=C2a=C2 the value g′′>0g″>0 indicating point of minima.

The minimum value is therefore C+2C

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