Concept:
For any positive real numbers a1, a2, a3, ….., an. The following relation holds:
A.M. ≥ G.M.
G.M. = \(\sqrt[n]{a_{1}a_{2}a_{3}.....a_{n}}\) ----(2)
Calculation:
For the maximum value of a^{3}b, there should be four terms among which three terms are containing variable a and one term containing the variable b. So that, when we multiply the terms, we get a3b. We have to do all this without distorting the equation given in the problems.
⇒ The terms of the equation would be a, a, a, and 3b.
Since A.M. ≥ G.M.
\(⇒ \frac{a+a+a+3b}{4}≥ \sqrt[4]{a\times a\times a\times 3b}\) [using (1) & (2)]
⇒ (3a + 3b)/4 ≥ (3a^{3}b)^{1/4}
⇒ [(3a + 3b)/4]^{4} ≥ 3a3b
⇒ (3/4)^{4} (a + b)^{4} ≥ 3a3b
⇒ (81/256) (a + b)4 ≥ 3a3b
⇒ 27/256 ≥ \(\dfrac{a^3b}{(a + b)^4}\)
Hence, the maximum value of \(\dfrac{a^3b}{(a + b)^4}\) is \(\dfrac{27}{256}\)