$Q.$ If $x\in\mathbb R$ and $(5^{1+x}+5^{1-x}),\frac{a}{2},(25^x+25^{-x})$ are in $AP$ then $a$ lies in interval of $?$
Attempt 1 : lets say that its an increasing AP then, $$5^{1+x}+5^{1-x}<\frac{a}{2}<25^x+25^{-x}$$
Mininmum value of $5^{1+x}+5^{1-x} $ & $ 25^x+25^{-x}$ will be (by AM-GM) , $10$ and $2$ .
and $$2<\frac{a}{2}<10$$
Hence interval of $a$ will be $[12,\infty)$
Attempt 2 : if $x,y,z$ are in AP then $$x+z=2y$$ By that $$a=5^{1+x}+5^{1-x}+25^x+25^{-x}$$
By AM-GM I can say $$\frac{5^{1+x}+5^{1-x}+25^x+25^{-x}}{4}>\sqrt[4]{5^{1+x}\times5^{1-x}\times25^x\times25^{-x}}$$ OR $$a>4\sqrt[4]{25}$$
Doubt : in my second attempt , why I didn't my answer matched the answer from first attempt .
Is there another way of solving this ?
The question is essentially to find the range of $ 25^x + 5^{1+x} + 5 ^ { 1 - x } + 25^{-x}$ subject to $ x \in \mathbb{R}$.
Using the substitution $ 5^x = y$, we want to find the range of $ y^2 + 5y + \frac{5}{y} + \frac{ 1}{y^2} $, subject to $ y > 0$.
Applying AM-GM creatively (and I'm leaving this step to you), show that the minimum occurs at
Hence $ a \in \ldots $.
Note: With reference to my comment, as it turns out equality happens for the same $x$ value for the LHS and RHS.