Question

If $x,y$ and $z$ are all positive, then the minimum value of $f(x,y,z)=x^3+12\left(\frac{yz}{x}\right)+16{\left(\frac{1}{yz}\right)}^{3/2}$ is

• A. $14$
• B. $24$
• C. $42$
• D. $3$

Answer 1

The correct option is B $24$

Let $x^3,4\left(\frac{yz}{x}\right),4\left(\frac{yz}{x}\right),4\left(\frac{yz}{x}\right),8{\left(\frac{1}{yz}\right)}^{3/2},8{\left(\frac{1}{yz}\right)}^{3/2}$
be six positive numbers.
Applying $A.M.\ge G.M.,$
$\Rightarrow \frac{f\left(x\right)}{6}\ge {[x^3\cdot 4\left(\frac{yz}{x}\right)\cdot 4\left(\frac{yz}{x}\right)\cdot 4\left(\frac{yz}{x}\right)\cdot 8{\left(\frac{1}{yz}\right)}^{3/2}\cdot 8{\left(\frac{1}{yz}\right)}^{3/2}]}^{1/6}$
$\Rightarrow f\left(x\right)\ge 24$
Minimum value of $f\left(x\right)$ is $24.$