1
$\begingroup$

Problem: Suppose that $x,y$ and $z$ are positive real numbers verifying $xy+yz+zx=1$ and $k,l$ are two positive real constants. The minimum value of the expression: $$kx^2+ly^2+z^2$$ is $2t_0$, where $t_0$ is the unique root of the equation $2t^3+(k+l+1)t-kl=0.$

My Attempt: Let $k=a+(k-a)$ and $l=b+(l-b)$, where $a,b\in \mathbb{R^+}$ and $a\leq k$ and $b\leq l.$ Then from the following Inequalities: $$ax^2+by^2\geq2\sqrt{ab}xy$$ $$(l-b)y^2+z^2/2\geq\sqrt{2(l-b)}yz$$ $$z^2/2+(k-a)x^2\geq\sqrt{2(k-a)}zx$$ We can deduce that $$kx^2+ly^2+z^2\geq2\sqrt{ab}xy+\sqrt{2(l-b)}yz+\sqrt{2(k-a)}zx.$$ Since $xy+yz+zx=1\implies 2\sqrt{ab}=\sqrt{2(l-b)}=\sqrt{2(k-a)}.$ After this stage I get two quadratic expression for $a$ and $b$, which seems far of from the main result. Where am I going wrong?

PS.Please do not use Calculus to solve this Problem.

$\endgroup$
  • $\begingroup$ In the 3-dimensional space, the points $(x, y, z)$ satisfying $$ k x^2 + l y^2 + z^2 = \mbox{some constant $c$} $$ constitute an ellipsoid (call it $E(c)$) with principal semi-axes of lengths $1/\sqrt{k}, 1/\sqrt{l}, 1$. The points satisfying $xy+yz+zx=1$ constitute a quadratic surface, which we will call $S$. Your problem, therefore, is a problem in 3-D geometry: find the smallest $c$ for which the ellipsoid $E(c)$ still intersects surface $S$. Surface $S$ is away from the origin, so $E(c)$ and $S$ intersect at a point at which they are tangent to each other. $\endgroup$ – user8960 Oct 27 '16 at 19:33
0
$\begingroup$

We need to find a maximal value of $m$, for which the inequality $$kx^2+ly^2+z^2\geq m(xy+xz+yz)$$ is true for all reals $x$, $y$ and $z$ or $$z^2-m(x+y)z+kx^2+ly^2-mxy\geq0$$ for which we need $m^2(x+y)^2-4(kx^2+ly^2-mxy)\leq0$ or

$(4k-m^2)x^2-2(2m+m^2)xy+(4l-m^2)y^2\geq0$ for all reals $x$ and $y$, for which we need

$(2m+m^2)^2-(4k-m^2)(4l-m^2)\leq0$ and $4k-m^2>0$, which gives

$m^3+(1+k+l)m^2-4kl\leq0$ and since for $m=2\sqrt{k}$ we obtain

$m^3+(1+k+l)m^2-4kl=8k\sqrt{k}+(1+k+l)4k-4kl>0$,

we see that the minimal value of $kx^2+ly^2+z^2$ is a positive root of the equation $$m^3+(1+k+l)m^2-4kl=0$$

which not exactly that you wish.

| cite | improve this answer | |
$\endgroup$

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.