2
$\begingroup$

Let $k\in\mathbb{N}$ and $x,y$ and $z$ are positive real number such that $x+y+z=1$. How can we find minimum of $f(x,y,z)$ where $$f(x,y,z)=\frac{x^{k+2}}{x^{k+1}+y^{k}+z^{k}}+\frac{y^{k+2}}{y^{k+1}+z^{k}+x^{k}}+\frac{z^{k+2}}{z^{k+1}+x^{k}+y^{k}}$$ I think (if I am right) that we can use chebyshev inequality but I am not sure.

Chebyshev Inequality

If $\{a_i\}_{i=1}^{n}$ and $\{b_i\}_{i=1}^{n}$ be two non-increasing sequences then $$n(a_1b_1+a_2b_2+\cdots+a_nb_n)\ge(a_1+a_2+\cdots+a_n)(b_1+b_2+\cdots+b_n)$$ Or if $\{a_i\}_{i=1}^{n}$ be non-increasing sequence and $\{b_i\}_{i=1}^{n}$ be a non-decreasing sequence then $$n(a_1b_1+a_2b_2+\cdots+a_nb_n)\le(a_1+a_2+\cdots+a_n)(b_1+b_2+\cdots+b_n)$$

Thanks for your hints, references or answers.

$\endgroup$
3
$\begingroup$

This is not the complete solution. In here, I give the range of values in which $f(x,y,z)$ lies. First, note that since $x\geq 0,~y\geq 0,~z\geq 0$ and $x+y+z=1$, we must have $0\leq x,y,z\leq 1$. Thus, for any $k\in \mathbb{N}$, we have \begin{eqnarray} x^{k+1}< x^{k},~y^{k+1}< y^{k},~\text{and}~z^{k+1}< z^{k}. \end{eqnarray}

Thus, we have \begin{align} f(x,y,z)&> \frac{x^{k+2}+y^{k+2}+z^{k+2}}{x^{k}+y^{k}+z^{k}}. \end{align}

Without loss of generality, assume $x\geq y\geq z$. Then, applying the first Chebyshev's inequality with $a_{1}=x^{2},~a_{2}=y^{2},~a_{3}=z^{2},~b_{1}=x^{k},~b_{2}=y^{k},~b_{3}=z^{k}$, we get \begin{align} 3(x^{k+2}+y^{k+2}+z^{k+2})&\geq ({x^{2}+y^{2}+z^{2}})({x^{k}+y^{k}+z^{k}}). \end{align}

We thus have \begin{align} f(x,y,z)&> \frac{1}{3}(x^{2}+y^{2}+z^{2})\\ &\stackrel{(a)}{\geq}\frac{1}{9}, \end{align} where, in writing (a) above, I have used the solution $(x^{*},y^{*},z^{*})=(1/3,1/3,1/3)$ to the following optimization problem: \begin{eqnarray} (x^{*},y^{*},z^{*})=\arg\min ~~(x^{2}+y^{2}+z^{2})\\ \text{subject to}~~~~~x+y+z=1,~x\geq 0,~y\geq 0,~z\geq 0. \end{eqnarray}

Thus, $f(x,y,z)>\frac{1}{9}$, and by inspection, we find $f(x,y,z)\leq 1$. Thus, $f(x,y,z)\in (\frac{1}{9},1]$.

$\endgroup$
  • $\begingroup$ Note $\,x,y,z>0$ $\endgroup$ – Behrouz Maleki Jul 11 '16 at 11:15
  • $\begingroup$ @BehrouzMaleki I have only hinted at a possible range for the values of $f(x,y,z)$. My range could be weaker than the actual range of values, and it might so happen that there exists no value of the triplet $(x,y,z)$ such that the equation you have given has a solution. $\endgroup$ – PN Karthik Jul 11 '16 at 12:49
  • $\begingroup$ This seems to follow from the arithmic-geometric inequality. The min is obtained when all three terms are equal, but there is only one solution, (1/3, 1/3, 1/3). $\endgroup$ – James Jul 11 '16 at 18:36

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.