I came across the below question in an IMO preparatory book and tried to solve it, I followed a certain steps and got a solution, but the book followed a slightly different approach and arrived at a totally different solution, I couldn't find any fault with both approaches.
The problems is as follows:
Find the least value of $a$ for which $5^{1+x}+5^{1-x}$, $a/2$, $25^x+25^{-x}$ are three consecutive terms of an Arithmetic Progression.
My approach :
since the three terms are to be in AP
$a/2 = (5^{1+x}+5^{1-x}+25^x+25^{-x})/2$
$\Rightarrow a = 5\cdot5^x+5\cdot5^{-x}+5^{2x}+5^{-2x}$. Let $5^x = t$ (Step A)
Then $a = 5t+(5/t)+t^2+t^{-2}$. Now let $u=t+1/t$
$\Rightarrow a = 5u+u^2-2$, since $u^2 = t^2+t^{-2}+2$ (Step B)
This is a quadratic $u^2+5u-(2+a) = 0$
for $u$ to be real $25+4(2+a) \ge 0$,
$\Rightarrow a \ge -33/4$.
I even tried maximizing the expression on the RHS in Step B , i.e. maximizing $u^2+5u-2$, then differentiating and equating to zero $2u+5 = 0$,
$\Rightarrow u = -5/2$, substituting this value in the expression
$(-5/2)^2+5(-5/2)-2 = -33/4$ Same solution as above
Approach followed in the book below.
It's the same up to step A mentioned above.
then the equation becomes
$a = (t-(1/t))^2 +5(\sqrt{t}-1/\sqrt{t}))^2 +12$
since there are two square expressions, the value of this expression is always $\ge12$, hence $a\ge12$.
Both approaches seem fine to me, can someone explain the ambiguity?
The expression $u=t+t^{-1}$ cannot take any value. First of all, $t=5^x>0$; the minimum of $u=t+1/t$ is taken at $t=1$ (AM-GM); hence $t+1/t\ge2$. All values $\ge2$ for $u$ can be obtained.
You then have $$ a=u^2+5u-2\ge 4+10-2=12 $$