The minimum value of the expression $\dfrac{(a^4+1)(b^4+1)(c^4+1)}{ab^2c}$ occurs when $a = b = c = 1$. This results in a value of 8. The discussion highlights the importance of symmetry in optimization problems involving real numbers and suggests that setting variables equal can often lead to finding extrema efficiently.
PREREQUISITESMathematicians, students studying calculus, and anyone interested in optimization problems involving real numbers.
jacks said:My Solution:
Using $\bf{A.M\geq G.M}$ separately
$\displaystyle \frac{a^4+1}{a} = a^3+\frac{1}{a} = a^3+\frac{1}{3a}+\frac{1}{3a}+\frac{1}{3a}$
So $\displaystyle a^3+\frac{1}{3a}+\frac{1}{3a}+\frac{1}{3a}\geq 4\sqrt[4]{a^3\cdot \frac{1}{3a}\cdot \frac{1}{3a}\cdot\frac{1}{3a}} = \frac{4}{\sqrt[4]{27}}.....(1)$
and equality hold when $\displaystyle a^3 = \frac{1}{3a} = \frac{1}{3a} = \frac{1}{3a},$ So $\displaystyle a = \frac{1}{\sqrt[4]{3}}>0$
Similarly $\displaystyle \frac{c^4+1}{c} = c^3+\frac{1}{c} = c^3+\frac{1}{3c}+\frac{1}{3c}+\frac{1}{3c}$
So $\displaystyle c^3+\frac{1}{3c}+\frac{1}{3c}+\frac{1}{3c}\geq 4\sqrt[4]{c^3\cdot \frac{1}{3c}\cdot \frac{1}{3c}\cdot\frac{1}{3c}} = \frac{4}{\sqrt[4]{27}}......(2)$
and equality hold when $\displaystyle c^3 = \frac{1}{3c} = \frac{1}{3c} = \frac{1}{3c},$ So $\displaystyle c = \frac{1}{\sqrt[4]{3}}>0$
Now for $\displaystyle \frac{b^4+1}{b^2} = b^2+\frac{1}{b^2}......(3)$
So $\displaystyle b^2+\frac{1}{b^2}\geq 2\sqrt{b^2\cdot \frac{1}{b^2}} = 2$
and equality hold when $\displaystyle b^2 = \frac{1}{b^2}\Rightarrow b = 1>0$
So Minimum value of $\displaystyle \frac{(a^4+1)(b^4+1)(c^4+1)}{ab^2c} = \frac{4}{\sqrt[4]{27}}\cdot 2 \cdot \frac{4}{\sqrt[4]{27}}=\frac{32}{3\sqrt{3}}$
Which is occur at $\displaystyle a = \frac{1}{\sqrt[4]{3}}\;,b = 1\;,c = \frac{1}{\sqrt[4]{3}}$