If $x+y+z=5$ and $xy+yz+zx=3$,then least and largest value of $x$ are?
Solution 1
Using $(x+y+z)^2=x^2+y^2+z^2+2(xy+yz+zx)$
So $x^2+y^2+z^2 = (x+y+z)^22(xy+yz+zx) = 5^22\cdot 3 = 19$
So $y^2+z^2 = 19x^2$ and $y+z = 5x$
Now Using $\bf{Cauchy\; Schwarz}$ Inequality
$$(y^2+z^2)(1^2+1^2)\geq (y+z)^2$$
So $$(19x^2)\cdot 2 \geq (5x)^2$$
So $$25+x^210x\leq 382x^2$$
So $$3x^210x13\leq 0$$
So $$3x^213x+3x13\leq0$$
So $$(x+1)(3x13)\leq 0\Rightarrow 1 \leq x\leq \frac{13}{3}$$
Solution 2
$x(y+z) = 3  yz \implies x(5x) = 3  yz \implies yz = x^25x+3\implies x^25x+3 \le \dfrac{(y+z)^2}{4}= \dfrac{(5x)^2}{4}\implies 4x^220x + 12 \le 2510x+x^2\implies 3x^210x13 \le 0 \implies (x+1)(3x13) \le 0 \implies 1 \le x \le \dfrac{13}{3} \implies x_{\text{max}} = \dfrac{13}{3}, x_{\text{min}} = 1.$
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Comments

Pole_Star 22 days
If $x+y+z=5$ and $xy+yz+zx=3$,then least and largest value of $x$ are?

dxiv about 6 yearsA hint at least of what you tried and where you got stuck, please?
