Find the minimum value of u, where $xyz = k^3$ and $u = (x + a)(y + b)(z + c)$ 
Find the minimum value of u, where
a) $$x^2 + y^2 = 1$$ and $$u = \frac{(ax^2 + by^2)}{\sqrt{(a^2x^2 + b^2y^2)}}$$
b) $$xyz = k^3$$ and $$u = (x + a)(y + b)(z + c)$$ and $$a > 0; b > 0; c > 0$$

I could do the first one with Lagrange  multiplier but it's really tedious. Can anyone give me any better way to solve the second one?? I don't want to use another tedious calculation with Lagrange multipliers. I've asked this same question few months ago but I couldn't work with the Holder inequality hint and also I failed to solve the equations arising using Lagrange multipliers.
 A: Lagrange multipliers is not hard for second case:
$$\frac{d}{dx}(u-\lambda(xyz-k^3))=(y+b)(z+c)-\lambda yz=0$$
$$(y+b)(z+c)=\lambda yz$$
Similarly, for the other variables, you get $$(x+a)(z+c)=\lambda xz\\(x+a)(y+b)=\lambda yz$$
Multiplying these last three equations by $x+a$, $y+b$ and $z+c$ respectively, you get $$yz(x+a)=xz(y+b)=xy(z+c)$$or $$ayz=bxz=cxy$$
You can write $y$ and $z$ in terms of $x$:
$$y=\frac ba x\\z=\frac ca x$$
Then plug those into your constraint:
$$x\frac ba x\frac ca x=k^3$$
$$x=k\sqrt[3]{\frac{a^2}{bc}}$$
Then the maximum value for $u$ is $$(k\sqrt[3]{\frac{a^2}{bc}}+a)(k\sqrt[3]{\frac{b^2}{ac}}+b)(k\sqrt[3]{\frac{c^2}{ab}}+c)=abc\left(\frac{k}{\sqrt[3]{abc}}+1\right)^3$$
A: Lagrange multipliers aren't really that bad here. We get
$$
(y+b)(z+c) = \lambda yz\\
(x+a)(z+c) = \lambda xz\\
(x+a)(y+b) = \lambda xy\\
xyz = k^3.
$$
Divide the first three equations pairwise and simplify to get
$$
\frac{x}{a} = \frac{y}{b}=\frac{z}{c}.
$$
Call this common ratio $r$. Plugging this into the constraint gives $r = k/\sqrt[3]{abc}$, and the minimum value is thus $(\sqrt[3]{abc}+k)^3$.
