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 $x^2+y^2+z^2 = (x+y+z)^2-2(xy+yz+zx) = 5^2-2\cdot 3 = 19$

So $y^2+z^2 = 19-x^2$ and $y+z = 5-x$

Now Using $\bf{Cauchy\; Schwarz}$ Inequality

$$(y^2+z^2)(1^2+1^2)\geq (y+z)^2$$

So $$(19-x^2)\cdot 2 \geq (5-x)^2$$

So $$25+x^2-10x\leq 38-2x^2$$

So $$3x^2-10x-13\leq 0$$

So $$3x^2-13x+3x-13\leq0$$

So $$(x+1)(3x-13)\leq 0\Rightarrow -1 \leq x\leq \frac{13}{3}$$

Solution 2

$x(y+z) = 3 - yz \implies x(5-x) = 3 - yz \implies yz = x^2-5x+3\implies x^2-5x+3 \le \dfrac{(y+z)^2}{4}= \dfrac{(5-x)^2}{4}\implies 4x^2-20x + 12 \le 25-10x+x^2\implies 3x^2-10x-13 \le 0 \implies (x+1)(3x-13) \le 0 \implies -1 \le x \le \dfrac{13}{3} \implies x_{\text{max}} = \dfrac{13}{3}, x_{\text{min}} = -1.$

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Updated on December 05, 2022