Find the minimum value of $u=(x+a)(y+b)(z+c)$ where $a,b,c>0$. Find the minimum value of $u=(x+a)(y+b)(z+c)$ where $a,b,c>0$. Given $xyz=k^3$.
I've tried simplifying the factors and use the relation $xyz=k^3$ to reduce the variables but I'm stuck, I have no clue where to go from here.
Can anyone guide me to solve this??
 A: Hint: Holders inequality extends to give
$$(x+a)(y+b)(z+c)\geqslant (k+\sqrt[3]{abc})^3$$
A: I'm going to assume you meant $u=(x+a)(y+b)(z+c)$? In any case, this is a nice Lagrange multiplier (see Wikipedia for reference) problem. So we have a function $u$ and a constraint $xyz=k^3$. Thus $\mathcal{L}(x,y,z,\lambda)=(x+a)(y+b)(z+c)-\lambda (xyz-k^3)$. Computing the gradient
$\nabla \mathcal{L}=[(b+y)(c+z)-\lambda yz,(a+x)(c+z)-\lambda xz,(a+x)(b+y)-\lambda xy,k^3-xyz]^T$
Now you from the condition $\nabla \mathcal{L}=0$ you get four equations to solve for the variables $x$,$y$,$z$, and $\lambda$. Hope this helps.
Edit: (Solving the equations)
Multiplying each of the first three equations by the position variables, so that we have $x \partial_x\mathcal{L}=0$ etc. and using the fourth equation $x y z=k^3$ we get
$$
I:b c x +c x y+b xz +k^3(1-\lambda) =0 \\
II:a c y + c x y + a y z +k^3(1-\lambda) =0 \\
III:a b z + b x z + a y z +k^3(1-\lambda) =0 \\
$$
writing the first to equations as
$$
I:c x y + b x (c + z) +k^3(1-\lambda) =0 \\
II:c x y + a y (c + z) +k^3(1-\lambda) =0 \\
$$
we get the condition $(bx-ay)(c+z)=0$. From there you have to do two things, You can get more conditions like this by rewriting equations I and III, II and III. Then follow the path $z=-c$ or $bx=ay$ and see what answers you get. Note that you should not be astonished that you get more than one answer, since the problem is symmetric in permutations of $a,b,c$ simultaneously with $x,y,z$
