<s>[INST] If $a,b,c>0$ and $a+b+c=1$, prove $\frac{a}{a+bc}+\frac{b}{b+ca}+\frac{\sqrt{abc}}{a+ba}\le 1+\frac{3\sqrt{3}}{4}$ If $a,b,c>0$ and $a+b+c=1$, then prove
$$\frac{a}{a+bc}+\frac{b}{b+ca}+\frac{\sqrt{abc}}{a+ba}\le 1+\frac{3\sqrt{3}}{4}$$
 [/INST] HINT: set $a=xy,b=yz,c=zx$ and after this use the $\tan(\alpha/2)$ etc substitution4}
after simplification we get 
$$\frac{1}{1+z^2}+ \frac {1}{1+x^2}+\frac {y}{1+y^2}\le 1+\frac{3\sqrt{3}}{4}$$
now set $$x=\tan(\alpha/2),y=\tan(\beta/2),z=\tan(\gamma/2)$$
after simplification we get
$$\cos(\gamma)+\cos(\alpha)+\sin(\beta)\le \frac{3\sqrt{3}}{2}$$
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