Prove that $\frac{1}{\sqrt{ab+a+2}}+ \frac{1}{\sqrt{bc+b+2}}+ \frac{1}{\sqrt{ac+c+2}} \leq \frac{3}{2}$

Question

Let $abc=1$ and $a,b,c>0$. Prove that $$\frac {1}{\sqrt {ab+a+2}}+ \frac {1}{\sqrt {bc+b+2}}+ \frac {1}{\sqrt {ac+c+2}} \leq \frac {3}{2}.$$

I guess that $$\frac {1}{\sqrt {ab+a+2}}+ \frac {1}{\sqrt {bc+b+2}}+ \frac {1}{\sqrt {ac+c+2}} = \frac {3}{2}$$ if and only if $a=b=c=1$. I have spent two nights on this but it's just a mess and I still don't know how to apply Cauchy Schwarz or something to write a rigorous proof for this exercise.

Answer 1

$\mathbf {Hint: }$ Let $$\frac{1}{\sqrt{ab+a+2}}+ \frac{1}{\sqrt{bc+b+2}}+ \frac{1}{\sqrt{ac+c+2}}=A$$ Applying C-S: $$\left(\frac {1}{ab+a+2}+\frac {1}{bc+b+2}+\frac {1}{ac+c+2}\right)(3) \ge (A)^2$$ Hence $$A\le \sqrt{\left(\frac {1}{ab+a+2}+\frac {1}{bc+b+2}+\frac {1}{ac+c+2}\right)(3)}$$

Now consider by Titu's lemma we have $$\frac {1}{ab+1} + \frac {1}{a+1} \ge \frac {4}{ab+a+2}$$ Hence our obtained inequality now has $$A\le \sqrt{ \left( \frac {1}{ab+a+2}+\frac {1}{bc+b+2}+\frac {1}{ac+c+2} \right) (3) }\le \sqrt { \frac{3}{4} \sum_{cyc} \left(\frac {1}{ab+1} + \frac {1}{a+1} \right) }$$ But $$\sum_{cyc} \left(\frac {1}{ab+1} + \frac {1}{a+1}\right) =\sum_{cyc} \left(\frac {c}{c+1} + \frac {1}{a+1}\right)=3 $$ Because $abc=1$ Hence $$A \le \sqrt {\frac{3}{4} \sum_{cyc} \frac {c}{c+1} + \frac {1}{a+1}}=\sqrt {\frac{9}{4}}=\frac {3}{2}$$

Hope it helped now.

Answer 2

Let $a=\frac{y}{x}$ and $b=\frac{z}{y}$, where $x$, $y$ and $z$ are positives.

Thus, $c=\frac{x}{z}$ and by C-S we obtain: $$\sum_{cyc}\frac{1}{\sqrt{ab+a+2}}=\sum_{cyc}\sqrt{\frac{1}{\frac{z}{x}+\frac{y}{x}+2}}=\sum_{cyc}\sqrt{\frac{x}{2x+y+z}}$$ $$\leq\sqrt{\sum_{cyc}1^2\sum_{cyc}\frac{x}{2x+y+z}}=\sqrt{3\sum_{cyc}x\cdot\frac{1}{x+z+x+y}}$$ $$\leq\sqrt{3\sum_{cyc}\frac{x}{(1+1)^2}\left(\frac{1^2}{x+z}+\frac{1^2}{x+y}\right)}=\sqrt{\frac{3}{4}\sum_{cyc}\left(\frac{x}{x+z}+\frac{x}{x+y}\right)}$$ $$=\sqrt{\frac{3}{4}\sum_{cyc}\left(\frac{y}{y+x}+\frac{x}{x+y}\right)}=\frac{3}{2}.$$