Find the minimum value of this Expression (Without Calculus!)

Question

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.

Answer 1

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.

Answer 2

Let $f = kx^2+ly^2+z^2,$ suppose $f_{\min} = 2t > 0.$ We need to prove $$kx^2+ly^2+z^2 \geqslant 2t(xy+xz+yz),$$ equivalent to $$kx^2+ly^2+z^2+t(x^2+y^2+z^2)\geqslant 2t(xy+xz+yz)+t(x^2+y^2+z^2),$$ or $$(k+t)x^2+(l+t)y^2+(1+t)z^2 \geqslant t(x+y+z)^2.$$ Using the Cauchy-Schwarz inequality, we have $$(k+t)x^2+(l+t)y^2+(1+t)z^2 \geqslant \frac{(x+y+z)^2}{\frac{1}{k+t}+\frac{1}{l+t}+\frac{1}{1+t}}.$$ Therefore, the constant $t$ must be satisfy $$\frac{1}{k+t}+\frac{1}{l+t}+\frac{1}{1+t} = \frac{1}{t},$$ or $$2t^3+(k+l+1)t-kl=0.$$ The proof is completed.

P/s. Many years ago I created this problem (here) but thought this was an old problem.