Given positives $a, b, c$ such that $a^2 + b^2 + c^2 + abc = 4$, prove that $$\large \frac{a}{\sqrt{(b^2 + 2)(c^2 + 2)}} + \frac{b}{\sqrt{(c^2 + 2)(a^2 + 2)}} + \frac{c}{\sqrt{(a^2 + 2)(b^2 + 2)}} \ge 1$$
We have that $$\sin^2\gamma + \cos^2\gamma = \sin^2\alpha + \cos^2\alpha = 1$$
$$\iff \cos^2\gamma(1 – \sin^2\alpha) = \cos^2\alpha(1 – \cos^2\gamma) = (\cos\gamma\cos\alpha)^2$$
$$\iff \cos^2\gamma + \cos^2\alpha – 2(\cos\gamma\cos\alpha)^2 = (\cos\gamma\sin\alpha)^2 + (\cos\alpha\sin\gamma)^2$$
$$\iff \cos^2\gamma + \cos^2\alpha – 2\cos\gamma\cos\alpha(\cos\gamma\cos\alpha – \sin\gamma\sin\alpha) = (\cos\gamma\sin\alpha +\cos\alpha\sin\gamma)^2$$
$$\iff \cos^2\gamma + \cos^2\alpha – 2\cos\gamma\cos\alpha\cos(\gamma + \alpha) = \sin^2(\gamma + \alpha)$$
$$\iff \cos^2\alpha + \cos^2\beta + \cos^2\gamma + 2\cos\alpha\cos\beta\cos\gamma = 1 \ (\beta = 180^\circ – \gamma – \alpha)$$
If we let $\dfrac{\cos\alpha}{a} = \dfrac{\cos\beta}{b} = \dfrac{\cos\gamma}{c} = 2$ then $a^2 + b^2 + c^2 + abc = 4$
It needs to be sufficient to prove that $$\frac{\cos\alpha}{\sqrt{(\cos^2\beta + 2)(\cos^2\gamma + 2)}} + \frac{\cos\beta}{\sqrt{(\cos^2\gamma + 2)(\cos^2\alpha + 2)}} + \frac{\cos\gamma}{\sqrt{(\cos^2\alpha + 2)(\cos^2\beta + 2)}} \ge 1$$
Another replacement is that $$\frac{\sqrt{\dfrac{yz}{(z + x)(x + y)}}}{a} = \frac{\sqrt{\dfrac{zx}{(x + y)(y + z)}}}{b} = \frac{\sqrt{\dfrac{xy}{(y + z)(z + x)}}}{c} = 2$$
We have that $$\sum_{cyc}\frac{zx}{(x + y)(y + z)} + \prod_{cyc}\sqrt{\frac{zx}{(x + y)(y + z)}} = \frac{\displaystyle\sum_{cyc}zx(z + x)}{(y + z)(z + x)(x + y)} + \prod_{cyc}\frac{y}{z + x} = 1$$
It needs to be sufficient to prove that $$\sum_{cyc}\sqrt\frac{zx(z + x)^2}{[2(y + z)(z + x) + xy][2(z + x)(x + y) + yz]} \ge 1$$
Neither of the above does I know how to solve.
Best Answer
I like the following way.
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.
Thus, $c=\frac{2z}{\sqrt{(x+z)(y+z)}}$ and we need to prove that: $$\sum_{cyc}\frac{\frac{2x}{\sqrt{(x+y)(x+z)}}}{\sqrt{\left(\frac{4y^2}{(x+y)(y+z)}+2\right)\left(\frac{4z^2}{(x+z)(y+z)}+2\right)}}\geq1$$ or $$\sum_{cyc}\frac{x(y+z)}{\sqrt{(3y^2+xy+xz+yz)(3z^2+xy+xz+yz)}}\geq1$$ or $$\sum_{cyc}x(y+z)\sqrt{3x^2+xy+xz+yz}\geq\sqrt{\prod_{cyc}(3x^2+xy+xz+yz)}$$ or $$\sum_{cyc}x^2(y+z)^2(3x^2+xy+xz+yz)+$$ $$+2\sum_{cyc}xy(x+z)(y+z)\sqrt{(3x^2+xy+xz+yz)(3y^2+xy+xz+yz)}\geq$$ $$\geq\prod_{cyc}(3x^2+xy+xz+yz).$$ But by C-S $$\sqrt{(3x^2+xy+xz+yz)(3y^2+xy+xz+yz)}\geq3xy+xy+xz+yz=4xy+xz+yz.$$ Id est, it's enough to prove that: $$\sum_{cyc}x^2(y+z)^2(3x^2+xy+xz+yz)+$$ $$+2\sum_{cyc}xy(x+z)(y+z)(4xy+xz+yz)\geq$$ $$\geq\prod_{cyc}(3x^2+xy+xz+yz),$$ which is identity!!!
Done!