computation of minimum value of a function

Question

How will we calculate the minimum value ilof the equation with the given condition.

Answer 1

Using Holder,s Inequality

$\displaystyle (x^2+8y^2+27z^2)\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)\geq \bigg[\bigg(x^2\cdot \frac{1}{x}\cdot \frac{1}{x}\bigg)^{\frac{1}{3}}+\bigg(8y^2\cdot \frac{1}{y}\cdot \frac{1}{y}\bigg)^{\frac{1}{3}}+\bigg(27z^2\cdot \frac{1}{z}\cdot \frac{1}{z}\bigg)^{\frac{1}{3}}\bigg]^{3}=(1+2+3)^3=216$

and equality hold when $\displaystyle x^3=8y^3=27z^3$

Answer 2

Hint: $$L(x,y,z,\lambda) = x^2+8y^2+27z^2+\lambda \left(1-\frac1x-\frac1y-\frac1z\right)$$

Differentiating with respect to $x$, we have $2x+\frac{\lambda}{x^2}=0$, $\lambda=-2x^3, $

Differentiating with respect to $y$, we have $16y+\frac{\lambda}{y^2}=0$, $\lambda=-16y^3$

Differentiating with respect to $z$, we have $54z+\frac{\lambda}{z^2}=0$, $\lambda=-54z^3$

Solve $x,y,z$ in terms of $\lambda$, then solve for $\lambda$ using $$\frac1x+\frac1y+\frac1z=1$$