Maximum value is not as big as you think

We just need to partition the terms in such a way that we can apply the AM-GM inequality .

Note that $\large a\sqrt [ 2 ]{ b } \sqrt [ 3 ]{ c } \sqrt [ 4 ]{ d } = \sqrt[12]{a^{12} b^6 c^4 d^3 }$ .

We split the terms as:
$36a = \underbrace{3a + 3a + \ldots + 3a}_{3a\text{'s added 12 times}}$ ,
$4b = \underbrace{\frac23b + \frac23b + \ldots+\frac23b}_{ \frac23\text{'s added 6 times}}$ ,
$4c = c + c + c + c + c$ , and
$3d = d + d + d$ .

Now we apply the inequality:

$\begin{aligned} \large \dfrac{36a+4b+4c+3d}{25} &\ge& \sqrt[\Large{25}]{ (3a)^{12} \cdot \left(\frac23b\right)^6\cdot c^4\cdot d^3} \\ \large 1^{25/12} & \ge& \sqrt[\Large{12}]{ (3a)^{12} \cdot \left(\frac23b\right)^6\cdot c^4\cdot d^3} \\ \large 1 & \ge& \sqrt6 a\sqrt [ 2 ]{ b } \sqrt [ 3 ]{ c } \sqrt [ 4 ]{ d }\\ \large \frac1{\sqrt6} & \ge& a\sqrt [ 2 ]{ b } \sqrt [ 3 ]{ c } \sqrt [ 4 ]{ d }\\ \end{aligned}$

The maximum value of the desired expression is $\dfrac1{\sqrt6} \approx \boxed{0.408}$ , and it occurs when $3a = \dfrac23b = 4c = 3d$ or $a = \dfrac13, b = \dfrac32, c = d = 1$ .

Here is a calculus way of doing it.

First, what we want to maximise is the same as maximising $\displaystyle a^{12}b^{6}c^{4}d^{3}$ since the answer is going to be the positive twelveth root of our expression.

So, our constrained Lagrangian is $\displaystyle L =a^{12}b^{6}c^{4}d^{3}-\lambda (36a+4b+4c+3d-25)$ .

Differentiating w.r.t. a,b,c and d and equating to 0, we get

$\displaystyle 0=12a^{11}b^{6}c^{4}d^{3}-36\lambda$

$\displaystyle 0=6a^{12}b^{5}c^{4}d^{3}-4\lambda$

$\displaystyle 0=4a^{12}b^{6}c^{3}d^{3}-4\lambda$

$\displaystyle 0=3a^{12}b^{6}c^{4}d^{2}-3\lambda$

which implies(Note that $\lambda$ is non zero since all a,b,c and d are.)

$\displaystyle b=\dfrac{9}{2}a, c=d=3a$ plugging this in the constraint, we have $\displaystyle a=\dfrac{1}{3}, b=\dfrac{3}{2}, c=d=1$

So, $\displaystyle a\sqrt{b}\sqrt[3]{c}\sqrt[4]{d}=\dfrac{1}{3}\sqrt{\dfrac{3}{2}}=\boxed{\sqrt{\dfrac{1}{6}}}$