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proving a property of inequality

for all $a,b,cin mathbb R^+$ such that $a+b+c=1$ and for all $x,y,zin mathbb R^+$
$$(x+y+z)^2leq xy+xz+yz$$
Here are my approaches:
1.triangle inequality
$$(x+y+z)^2leq xy+xz+yz$$
if we factorize this we get
$$(x+y+z)(x+y+z)leq xy+xz+yz$$
now multiply both sides with $a+b+c=1$
$$(x+y+z)(a+b+c)leq xy+xz+yz$$
$$(a+b+c)(x+y+z)leq xy+xz+yz$$
proof is complete using AM-GM inequality
2.adding it up
$$x(a+b+c)+y(a+b+c)+z(a+b+c)leq xy+xz+yz+3abc$$
now we have $a+b+c=1$ and after canceling of $x,y,z$
$$3abc+3abc=3abc$$
what should I do next?

A:

As you have pointed out:
$$x(a+b+c) +y(a+b+c) + z(a+b+c) = (x+y+z)(a+b+c)$$
Rearrange it to form:
$$(a+b+c)(x+y+z) leq xy+x
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