MHB Find Least Value of a,b,c Real Nums

[anemone]

Find the least value of $\dfrac{(a^4+1)(b^4+1)(c^4+1)}{ab^2c}$ as $a,\,b,\,c$ range over the positive reals.

 

[juantheron]

My Solution:

Spoiler

Using $\bf{A.M\geq G.M}$ Seperately

$\displaystyle \frac{a^4+1}{a} = a^3+\frac{1}{a} = a^3+\frac{1}{3a}+\frac{1}{3a}+\frac{1}{3a}$

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 $\displaystyle a = \frac{1}{\sqrt[4]{3}}\;,b = 1\;,c = \frac{1}{\sqrt[4]{3}}$

 

[anemone]

My Solution:

Spoiler

Using $\bf{A.M\geq G.M}$ Seperately

$\displaystyle \frac{a^4+1}{a} = a^3+\frac{1}{a} = a^3+\frac{1}{3a}+\frac{1}{3a}+\frac{1}{3a}$

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 $\displaystyle a = \frac{1}{\sqrt[4]{3}}\;,b = 1\;,c = \frac{1}{\sqrt[4]{3}}$

Well done, jacks! Thanks for participating!