$\sqrt{ab}$ + $\sqrt{bc}$ + $\sqrt{ac}$ $\leq$ $ \frac{3}{2}$

Question

If $ a, b, c \in R^{+}$ and $\frac{1}{a + 1}$ + $\frac{1}{b + 1}$ + $\frac{1}{c + 1}$ = 2, then $\sqrt{ab}$ + $\sqrt{bc}$ + $\sqrt{ac}$ $\leq$ $ \frac{3}{2}$.

I don't know how to start.

Answer 1

the condition is equivalent to $$ab+ac+bc+2abc=1$$ setting $$\sqrt{ab}=x,\sqrt{bc}=y,\sqrt{ac}=z$$ and we get $$xyz==abc$$ and we have to show that $$x+y+z\le \frac{3}{2}$$ the last step is setting $$x=\cos(\alpha),y=\cos(\beta),z=\cos(\gamma)$$ Now we use the wellknown fact that $$\cos(\alpha)^2+\cos(\beta)^2+\cos(\gamma)^2+2\cos(\alpha)\cos(\beta)\cos(\gamma)=1$$ and $$\cos(\alpha)+\cos(\beta)+\cos(\gamma)=1+\frac{r}{R}$$

Answer 2

The condition gives $\sum\limits_{cyc}\left(\frac{1}{1+a}-1\right)=-1$ or $\sum\limits_{cyc}\frac{a}{1+a}=1$.

Thus, by C-S $1=\sum\limits_{cyc}\frac{a}{1+a}\geq\frac{\left(\sqrt{a}+\sqrt{b}+\sqrt{c}\right)^2}{3+a+b+c}$, which gives what you wanted.