Find the minimum value of u, where $xyz = k^3$ and $u = (x + a)(y + b)(z + c)$

Question

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.

Answer 1

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 $(\sqrt[3]{abc}+k)^3$.

Answer 2

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 $$(k\sqrt[3]{\frac{a^2}{bc}}+a)(k\sqrt[3]{\frac{b^2}{ac}}+b)(k\sqrt[3]{\frac{c^2}{ab}}+c)=abc\left(\frac{k}{\sqrt[3]{abc}}+1\right)^3$$