# If $x+y+z=5$ and $xy+yz+zx=3$,then least and largest value of $x$ are? [closed]

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

• A hint at least of what you tried and where you got stuck, please?
– dxiv
Dec 23, 2016 at 5:49

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

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

Rearranging the given equations a bit,

$(x + y)^2 = (5 - z)^2$
$xy = 3 - z(x + y) = 3 - z(5 - z)$

Now,

$0 ≤ (x - y)^2 = (x + y)^2 - 4xy$

We can substitute for both $x + y$ and $xy$ giving us an inequality involving only the variable $z$:

$0 ≤ (x + y)^2 - 4xy = 25 - 10z + z^2 - 12 + 20z - 4z^2 = -3z^2 + 10z + 13$.

Since this inequality holds for $z$ we can determine all possible values of $z$:

$0 ≤ -3z^2 + 10z + 13 = -(z + 1)(3z - 13)$

The inequality holds when $-1 ≤ z ≤ \frac{13}{3}$

Since the given equations for $x$, $y$, and $z$ can be manipulated to form this same quadratic inequality in $x$, $y$, or $z$ (basically using the symmetry of the equations involved), they each have maximum possible values of $\frac{13}{3}$. This maximum can be achieved when $x = y = \frac{1}{3}$ and $z = \frac{13}{3}$.

We can write the first equation as: $$(z + y)^2 = (5 - x)^2$$ and write the second one as, $$yz = 3 - x(z + y) = 3 - x(5 - x)$$

Now we also know that

$$0 \leq (z - y)^2 = (z + y)^2 - 4yz$$

Substituting we have:

$$0 \leq (z + y)^2 - 4yz = (25 - 10x + x^2) - 12 + 20x - 4x^2 = -3x^2 + 10x + 13 =-(x+1)(3x-13)$$

The inequality holds when $-1 \leq x \leq \frac{13}{3}$.

Thus the minimum value of $x$ is $-1$ and its maximum is $\frac{13}{3}$. Hope it helps.