A different way of asking questions

$Let\quad x=1/3+a,y=1/3+b,and\quad z=1/3+c.\\ x+y+z=1/3+a+1/3+b+1/3+c=1+a+b+c.\\ But\quad as\quad x+y+z=1,we\quad deduce\quad that\quad a+b+c=0.\\ \therefore { (a+b+c) }^{ 2 }={ a }^{ 2 }+{ b }^{ 2 }+{ c }^{ 2 }+2(ab+ac+bc)=0\\ \quad \quad 2(ab+ac+bc)=-({ a }^{ 2 }+{ b }^{ 2 }+{ c }^{ 2 })\\ \therefore \quad ab+ac+bc=-({ a }^{ 2 }+{ b }^{ 2 }+{ c }^{ 2 })/2=-d,where\quad d\ge 0\\ So\quad xy+xz+yz\\ =(1/3+a)(1/3+b)+(1/3+a)(1/3+c)+(1/3+b)(1/3+c)\\ =1/9+a/3+b/3+ab+1/9+a/3+c/3+ac+1/9+b/3+c/3+bc\\ =1/3+(2/3)(a+b+c)+ab+ac+bc\\ As\quad a+b+c=0\quad and\quad ab+ac+bc=-d,we\quad get,\\ S=xy+xz+yz=1/3-d1/3=1/3+(2/3)(a+b+c)+ab+ac+bc\\ As\quad a+b+c=0\quad and\quad ab+ac+bc=-d,we\quad get,\\ S=xy+xz+yz=1/3-d\le 1/3$

First of all, this was a nice question. I have taken an unconventional way to solve this problem. enjoy :D

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We Have: $x,y,z\in\mathbb{R}$ and $x+y+z=1$

We need to calculate the maximum value of $S=xy+yz+zx$

LET: $x,y,z$ be the three real roots of a monic cubic polynomial $P(a)$

By Vieta's Relations, we have:

$\large{P(a)=a^{3}-a^{2}+Sa+t}$ where $t$ is a constant.

we also have: $\large{P'(a)=3a^{2}-2a+S}$

Now for a cubic polynomial $P(a)$ to have 3 real roots it should change it's direction(i.e. the $\text{sign of it's slope}$ ) 3 times.

more importantly, thus, the curve of $P(a)$ should be $\textbf{parallel to the X-axis 2 times}$ ...

IMPLYING THAT: $P'(a)$ should be equal to $0$ twice!

So, it is clear that $P'(a)$ should have two real roots.

$\therefore \text{for } P'(a)=3a^{2}-2a+s$

$\large{\Delta=2^{2}-4.3.S≥0}$

$\implies \large{S≤\dfrac{1}{3}=\boxed{0.3}}$

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What do you think @abdulrahman khaled .. :D

We\quad know\quad that\quad for\quad real\quad numbers\quad x,y\quad and\quad z\quad { (x-y) }^{ 2 }+{ (x-z) }^{ 2 }+{ (y-z) }^{ 2 }\ge 0.\\ Therefore\quad { x }^{ 2 }+{ y }^{ 2 }+{ z }^{ 2 }\ge xy+yz+zx.So\quad { x }^{ 2 }+{ y }^{ 2 }+{ z }^{ 2 }+2xy+2yz+2zx\ge 3xy+3yz+3zx.Therefore\quad \\ { (x+y+z) }^{ 2 }\ge 3S.\gg \quad \frac { 1 }{ 3 } \ge S.So\quad the\quad final\quad answer\quad is\quad 0.3\quad

I really think that this is overrated.

${(x + y + z)}^{2} = {x}^{2} + {y}^{2} + {z}^{2} + 2(xy + yz + zx)$

$\frac{{(x + y + z)}^{2} - ({x}^{2} + {y}^{2} + {z}^{2})}{2} = xy + yz + zx$

Now we have to just find the minimum value of the LHS.

LHS = $\frac{1 - ({x}^{2} + {y}^{2} + {z}^{2})}{2}$

Using Cauchy Schwarz like this -

$({x}^{2} + {y}^{2} + {z}^{2})({1}^{2} + {1}^{2} + {1}^{2}) \geq {(x+y+z)}^{2}$

$({x}^{2} + {y}^{2} + {z}^{2}) \geq \frac{1}{3}$

LHS $\le \frac{1 - \frac{1}{3}}{2}$

$\le \frac{1}{3}$

Hence, $xy + yz + zx \le \frac{1}{3}$

x+y+z=1, pretend x=y=z=1/3, S=1/9+1/9+1/9 = 1/3 =0.333333333333333333.....

as ( x+y+z)^2 = x^2 + y^2 + z^2 + 2S so x^2 + y^2 + z^2 =1 - 2S Now (x-y)^2 + (y-z)^2 +(z-x)^2 = 2( x^2 + y^2 + z^2) - 2S i.e. 2(1-2S) - 2S >=0 or 1 - 3S >= 0 Hence S <= 1/3