Respected all. First of all, a very happy new year ( belated ) to all of you. I got stuck in the following problem. Please help me.

While I was solving problems on inequalities I came to know about the following problem:

Let $x,y,z$ be all positive real numbers such that $2x+3y+6z=k$ where $k$ is a constant. Find the minimum value of $x^2+y^2+z^2$.

I tried to solve the problem using the following

If $b_1, \cdots, b_n$ be positive real numbers, not all equal and $p_1, \cdots, p_n$ be their respective rational weights then for any rational $m$ we shall have $$\frac{p_1b_1^m+\cdots +p_nb_n^m}{p_1+\cdots+p_n}> or < \left(\frac{p_1b_1+\cdots+p_nb_n}{p_1+\cdots+p_n}\right)^m$$ according as $m$ does not or does lie within $(0,1).$

So by this, for $n=3$ we take $b_1=\frac x2, b_2=\frac y3, b_3=\frac z6, p_1=2^2, p_2=3^2, p_3=6^2, m=2$ we shall see

$$x^2+y^2+z^2>(\frac k7)^2$$

From this it came to my mind, what if we were asked to determine the minimum value of $x^a+y^a+z^a$ for some rational number $a$ other than 0,1 such that $lx+my+nz=k$ for some positive constant $k$ was given ? Using the above, can we derive the answer ? I tried to get the answer but in vain. Please help me, thanks in advance.

P.S. Please feel free to edit the question wherever you find error.

  • $\begingroup$ Geometric interpretation of the problem: $2x+3y+6z=k$ represents a plane, and $x^2+y^2+z^2$ represents the distance to the origin, squared. The minimum value is therefore the squared distance from the plane to the origin, which is $k^2/(2^2+3^2+6^2)$ $\endgroup$
    – f1garo
    Jan 13 '17 at 4:33
  • $\begingroup$ please answer the general question $\endgroup$
    – KON3
    Jan 13 '17 at 4:35
  • $\begingroup$ Maybe we can approach the problem geometrically. $z_\text{iso}^a = \Phi - x^a - y^a, \Phi \in \Bbb R^+$ are the isosurfaces of the cost function. We can imagine the plane $lx+my+nz=k$ being tangential to one of the isosurface with minimum cost. Differentiate $z_\text{iso}$ partially w.r.t. $x$ and $y$ to get the equations needed. $\endgroup$
    – f1garo
    Jan 13 '17 at 4:41
  • $\begingroup$ but what if I had to teach this in elementary way to graduate students ? I mean I want the answer using the above inequality only. Isn't it possible ? $\endgroup$
    – KON3
    Jan 13 '17 at 5:57

Case: $a> 1$.
Here we can use Holder's inequality in the form: $$\left(x^a+y^a+z^a\right)^{\frac1a} \cdot \left(l^{\frac{a}{a-1}}+ m^{\frac{a}{a-1}}+n^{\frac{a}{a-1}}\right)^{\frac{a-1}a} \geqslant k$$ $$\implies x^a+y^a+z^a \geqslant \frac{k^a}{\left(l^{\frac{a}{a-1}}+ m^{\frac{a}{a-1}}+n^{\frac{a}{a-1}}\right)^{a-1}}$$

Case: $0< a\leqslant 1$.
Now the objective function is concave in each of its arguments, so the minimum will be at one of the corners of the feasible region (which is convex). We may observe WLOG if $l = \max(l, m, n)$, then minimum is when $x=\frac{k}l,\; y=z=0$ to give a minimum value of $(k/l)^a$ or in general $$x^a+y^a+z^a \geqslant \left(\dfrac{k}{\max(l, m, n)}\right)^a$$

P.S. To specifically answer your question on whether your approach will work, for $a> 1$, any convexity / mean inequalities rooted approach should work. Specifically, for your weighted power mean approach you can manipulate the weights with $r = \frac{a}{a-1}>1$ like so: $$\frac{l^r\left(\frac{x}{l^{r-1}}\right)^a+m^r\left(\frac{y}{m^{r-1}}\right)^a+n^r\left(\frac{z}{n^{r-1}}\right)^a}{l^r + m^r + n^r} \geqslant \left(\frac{lx+my+nz}{l^r+m^r+n^r}\right)^a$$ which reduces to the expression in the first case above. For $a \in (0, 1]$ this approach fails as the minimum is not when the variables are all equal, but when they are on the corner of the feasible set.


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