Show that $a+b+c+\sqrt {\frac {a^2+b^2+c^2}{3}}\le4$, Let $a,b,c$ be positive real numbers such that $a^2+b^2+c^2+abc=4.$ Show that $$a+b+c+\sqrt {\frac {a^2+b^2+c^2}{3}}\le4.$$
 A: Let $f(x,y,z) = x+y+z + \sqrt{\dfrac{x^{2} + y^{2}+z^{2} }{3}}$ and $g(x,y,z) =  x+y+z + xyz$. By Lagrange's Method of Undetermined Multipliers, we will maximise $f$ subject to the constraint that $g=4$. (The set $D=\{(x,y,z): g(x,y,z)=4\}$ is closed and bounded and since $g$ is continuous, by the Extreme Value Theorem it attains its bounds on $D$.) 
By Lagrange's Method,
$\nabla f= \lambda \nabla g$
$\Rightarrow f_{x} = \lambda g_{x}$
$\Rightarrow f_{y} = \lambda g_{y}$
$\Rightarrow f_{z} = \lambda g_{z}$
$\Rightarrow 1 + \dfrac{2x}{\sqrt{3(x^{2} + y^{2}+z^{2})}} = \lambda(2x + yz)$ $\cdots (1)$
$\Rightarrow 1 + \dfrac{2y}{\sqrt{3(x^{2} + y^{2}+z^{2})}} = \lambda(2y + xz)$
$\Rightarrow 1 + \dfrac{2z}{\sqrt{3(x^{2} + y^{2}+z^{2})}} = \lambda(2z + yx)$
From the above $3$ equations, subtracting one the other yields the system below:
$(x-y)\left[\dfrac{2}{\sqrt{3(x^{2} + y^{2}+z^{2})}}-\lambda(2-z)\right] =0$
$(y-z)\left[\dfrac{2}{\sqrt{3(x^{2} + y^{2}+z^{2})}}-\lambda(2-x)\right] =0$
$(z-x)\left[\dfrac{2}{\sqrt{3(x^{2} + y^{2}+z^{2})}}-\lambda(2-y)\right] =0$
$\Rightarrow x=y=z$ or 
${\lambda}(z-2)={\lambda}(x-2)=\lambda(y-2)$
Clearly $\lambda \neq 0$ since if that were true, plugging back in $(1)$ would result in 
$\Rightarrow 1 + \dfrac{2x}{\sqrt{3(x^{2} + y^{2}+z^{2})}} =0$, which has no solutions when $x,y,z \in \mathbb{R}^{+}$.
Thus in both cases we see that $x=y=z=a$, which after plugging into $g(x,y,z)=4$ gives:
$a^{3}+3a^{2}-4 =0$
$\Rightarrow (a-1)(a+2)^{2}=0$
$\Rightarrow a=1$ since $a \in \mathbb{R}^{+}$
$\Rightarrow f$ has an extremum at $x=y=z=1$ at which $f(x,y,z)=4$. Clearly this is a point of maximisation since if it were a point of minimisation it would be contradicted by the fact that $f\left(\dfrac{1}{2}, \dfrac{1}{2},\dfrac{1}{2}\right) = \dfrac{7}{8} <4$. This proves the desired inequality.
A: Let $a=\frac{2x}{\sqrt{(x+y)(x+z)}}$ and $b=\frac{2y}{\sqrt{(x+y)(y+z)}},$ where $x$, $y$ and $z$ are positives.
Hence, $c=\frac{2z}{\sqrt{(x+z)(y+z)}}$ and we need to prove that
$$\sum_{cyc}\frac{2x}{\sqrt{(x+y)(x+z)}}+\sqrt{\frac{4}{3}\sum_{cyc}\frac{x^2}{(x+y)(x+z)}}\leq4$$ or
$$\sum_{cyc}x\sqrt{y+z}+\sqrt{\frac{1}{3}\sum_{cyc}(x^2y+x^2z)}\leq2\sqrt{(x+y)(x+z)(y+z)}.$$
Let $\sum\limits_{cyc}(x^2y+x^2z)=6kxyz$.
Hence, by C-S
$$\sum_{cyc}x\sqrt{y+z}\leq\sqrt{\sum_{cyc}x\sum_{cyc}x(y+z)}=\sqrt{2\sum_{cyc}(x^2y+x^2z+xyz)}.$$
Thus, it's enough to prove that
$$\sqrt{2(6k+3)}+\sqrt{\frac{1}{3}\cdot6k}\leq2\sqrt{6k+2}$$ or
$$\sqrt{6k+3}+\sqrt{k}\leq2\sqrt{3k+1},$$
which is C-S again:
$$\sqrt{6k+3}+\sqrt{k}=3\sqrt{\frac{2k+1}{3}}+\sqrt{k}\leq\sqrt{(3+1)\left(3\cdot\frac{2k+1}{3}+k\right)}=2\sqrt{3k+1}.$$
Done!
