<s>[INST] Find the minimum value of this Expression (Without Calculus!) 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. 
 [/INST] 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.
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