Inequality with 6 variables I am trying to show the following,
Let, $x,y,z,a,b,c$ be positive reals such that $x+y+z=1$
Then $ax + by + cz +2 \sqrt{(ab+bc+ca)(xy+yz+zx)} \leq a+b+c$
This is practice problem 6 in this document
This is equivalent to 
$\sqrt{(ab+bc+ca)(xy+yz+zx)} \leq \frac{(y+z)a+(x+z)b+(x+y)c}{2(x+y+z)}$
But I am not avle to apply Jensen's inequality. Thank you for help

Apply Cauchy-Schwarz inequality:
$ab+bc+ca \leq \dfrac{(a+b+c)^2}{3}$,
$xy+yz+zx \leq \dfrac{(x+y+z)^2}{3}$.
Thus: $LHS \leq ax+by+cz +\dfrac{2(a+b+c)(x+y+z)}{3} \leq x+y+z \iff 2(ax+by+cz) \leq a(y+z)+b(x+z)+c(x+y) \iff 3(ax+by+cz) \leq a+b+c \iff px+qy+rz \leq \dfrac{1}{3}, p+q+r=1, p = \dfrac{a}{a+b+c}, q = \dfrac{b}{a+b+c}, r = \dfrac{c}{a+b+c}$.
At this point, if you are okay with Lagrange Multiplier, then you are just "right there" at the answer, namely: Let $f(x,y,z) = px+qy+rz$, with $p+q+r = 1 = x+y+z$.
$f_x = \lambda = f_y =f_z \Rightarrow p = q = r = \dfrac{1}{3} \to f \leq \dfrac{1}{3}x+\dfrac{1}{3}y+\dfrac{1}{3}z = \dfrac{1}{3}(x+y+z) = \dfrac{1}{3}$.
