Find the minimum value of $2a+ (1/a) + (1/2b) + b$, where a, b > 0 Find the minimum value of $2a+ (1/a)  + (1/2b) + b$, where a, b > 0
My approach:-
Since a and b are positive numbers, I applied AM-GM inequality
$(2a+(1/a)+(1/2b)+b) /4$ ≥ $(2a\cdot (1/a)\cdot (1/2b)\cdot b) ^ {1/4} $
Giving me the answer as $4$
but the correct answer given in my textbook is $3\cdot \sqrt(2)$
I guess I am applying the inequality correctly, then what is the issue ?
 A: AM-GM inequality applied to $(2a,1/a,1/2b,b)$ gives
$$2a+\frac{1}{a}+\frac{1}{2b}+b \ge 4 \tag{1}$$
where equality is to be achieved for $2a=1/a=1/2b=b$. Solving this gives $a=1/\sqrt{2}$ but does not give a unique/consistent value for $b$.
Conclusion is, AM-GM cannot be applied to all four quantities at once. However, since $a,b$ are independent positive quantities, one can safely apply AM-GM separately to $(2a,1/a)$ and $(1/2b,b)$.
$$2a+\frac{1}{a}\ge 2\sqrt{2}$$
where equality is achieved for $2a=1/a \Rightarrow a=1/\sqrt{2}$. Similarly
$$\frac{1}{2b}+b\ge \sqrt{2}$$
where equality is achieved for $1/2b=b \Rightarrow b=1/\sqrt{2}$.
Hence $2a+1/a + 1/2b+b$ is minimized for $a=1/\sqrt{2}$ and $b=1/\sqrt{2}$. At these values, the expression takes the value of
$$2\sqrt{2}+\sqrt{2}=3\sqrt{2}$$
Indeed $3\sqrt{2}$ is larger than $4$ because our first inequation, $(1)$, still holds true.
A: In the way you are applying AM-GM, the equality condition is not achievable. If applying AM-GM to find minimum, please be sure to check the values at which equality occurs. In this case there are no such values.
Alternatively, complete the square and rewrite it as -
$ \small \displaystyle 2a + \frac 1 a + b + \frac 1 {2b} = \left(\sqrt{2a} - \frac{1}{\sqrt a}\right)^2 + \left(\sqrt b - \frac{1}{\sqrt{2b}}\right)^2 + 3 \sqrt2$
The minimum value of the first two terms can be zero which occurs at $a = b = \frac{1}{\sqrt 2}$
