2010/029) Find the maximum value of, (√p + √q + √r ). (plz see details. . . )?

Let α, β, γ > 0 and (α + β + γ) = π/2. If p = (tan α tan β) + 5, q = (tan β tan γ ) + 5
and r = (tan γ tan α) + 5 ,
find the maximum value of (√p + √q + √r ).

tan γ = tan [ (π/2) - (α+ß) ] = cot(α+ß) = 1 / tan(α+ß)

tan γ = (1 - tan α tan ß) / ( tan α + tan ß )

tan γ tan α + tan ß tan γ = 1 - tan α tan ß

(r-5) + (q-5) = 1 - (p-5)

p + q + r = 16 ................. (1)

Let p = a^, q= b^2 and r = c^2

We are given p+q+r = 16 => a^2+b^2+c^2 = 16

We need to maximize a + b + c

We know
(a+b+c)^2 = a^2+b^2 +c^2 + 2ab + 2bc + 2ca
(a-b)^2 = a^2 + b^2 – 2ab
(b-c)^2 = b^2 + c^2 – 2bc
(c-a)^2 = c^2 + a^2 – 2ac

Adding we get (a+b+c)^2 +(a-b)^2 + (b-c)^2 + (c-a)^2 = 2 (a^2+b^2 + c^2) = 32
Clearly a+b+c is maximum when a=b=c because as (a-b)^2 + (b-c)^2 + (c-a)^2 can not be lower than zero and is zero when a= b= c

So 2 (3a^2) = 32 or a^2 = p = 16/3

so maximum value of (√p + √q + √r ) = 4 √3