if $x+y+z=2$ and $xy+yz+zx=1$,Prove $x,y,z \in \left [0,\frac{4}{3} \right ]$

Question

if $x+y+z=2$ and $xy+yz+zx=1$,Prove $x,y,z \in \left [0,\frac{4}{3} \right ]$

things i have done: first thing to do is to show that $x,y,z$ are non-negative. $$xy+yz+zx=1 \Rightarrow zx=1-yz-xy \Rightarrow zx=1-y(z+x)\Rightarrow zx=1-y(2-y)=(y-1)^2$$

this equality shows that $x$ and $z$ have same sign(both of them are positive or negative).like this we can conclude that $xy=(x-1)^2$ and $yz=(x-1)^2$.So all variables are positive or negative.If all of them are negative then $x+y+z \neq 2$ so all of $x,y,z$ are non negative. I don't know what to do for showing that none of $x,y,z$ will be be bigger than $\frac{4}{3}$.

Answer 1

Consider the polynomial: $$ p(t)=(t-x)(t-y)(t-z) = t^3-2t^2+t+k = t(t-1)^2+k. $$ We know that $p(t)$ has three real roots, hence $-k=xyz$ is bounded between the two values of $t(t-1)^2$ in its stationary points. A stationary point obviously occurs for $t=1$, the other one occurs for $t=1/3$. The three reals roots of $p(t)$ hence belong to the interval $[0,u]$, where $u>1$ is the only real number such that $$ u(u-1)^2 = \frac{1}{3}\left(\frac{1}{3}-1\right)^2 = \frac{4}{27}. $$ Now it is straightforward to check that $u=\frac{4}{3}$.

For a visual proof:

$\quad$

Answer 2

Note we also have $x^2+y^2+z^2=2$ (expand $(x+y+z)^2=4$).

Note $y^2+z^2 \geq 2yz=2-2x(y+z)=2-2x(2-x)$, so

$x^2+y^2+z^2 \geq 3x^2-4x+2$

But $3x^2-4x+2 >2$ for $x>\frac{4}{3}$.

Answer 3

You have

$$ 1=xy+z(x+y)=xy+(x+y)(2-(x+y))=xy+x(2-x)+2y-y^2, $$

so that

$$ 0=y^2+(x-2)y+x^2-2x+1=\bigg(y+\frac{x-2}{2}\bigg)^2+\frac{3}{4}x^2-x $$

and hence $x-\frac{3}{4}x^2 =-\bigg(y+\frac{x-2}{2}\bigg)^2 \leq 0$. So we have $\frac{3x}{4}(\frac{4}{3}-x) \leq 0$, i.e. $x(x-\frac{4}{3}) \geq 0$, hence $x\in [0,\frac{3}{4}]$. By symmetry, the same holds for $y$ or $z$.

Answer 4

We can infact prove that, if $x \le y \le z$ then $x \in[0,1/3] ,y \in [1/3,1] , z \in [1/4/3] $

Consider the cubic whose roots are $x,y,z$ . Also let $xyz=k$ Then by veita relations, cubic will be $f(m)=m^3-2m^2+m+p$.

Now , $f'(m)=3m^2-4m+1=(3m-1)(m-1)$ So $f$ increases in interval $[-\infty,1/3]\cup[1,\infty] $

Notice that, $f(\frac{1}{3})=k+\frac{4}{27}$, $f(1)=k$ So $f$ will have three roots iff $k+\frac{4}{27} \ge 0, k \le 0 $ One more interesting thing is $f(0)=k, f(4/3)=\frac{4}{27}+k$. So there is one sign change between $x\in [0,\frac{1}{3}]$ $x \in [\frac{1}{3},1]$ and $x \in [1,\frac{4}{3}] $

So by continuity of polynomials, we deduce that there is root between each of these intervals. And this exactly what we wanted to prove $\Box$