How do I finish proving this inequality?

Question

Given $a, b,c >0$ satisfying $abc=1$. Prove that: $\dfrac{1}{{{a^2} + 2{b^2} + 3}} + \dfrac{1}{{{b^2} + 2{c^2} + 3}} + \dfrac{1}{{{c^2} + 2{a^2} + 3}} \le \dfrac{1}{2}$?

This is my try:

Using the AM–GM inequality, I get: $a^2+b^2 \ge 2ab$ and $b^2+1 \ge 2b$

Therefore, $a^2+2b^2+1 \ge 2ab+2b$ which implies that: $a^2+2b^2+3 \ge 2(ab+b+1)$

Hence, $\dfrac{1}{a^2+2b^2+3} \le \dfrac{1}{2(ab+b+1)}$

Doing the same, I get $\dfrac{1}{b^2+2c+3} \le \dfrac{1}{2(bc+c+1)}$ and $\dfrac{1}{c^2+2a+3} \le \dfrac{1}{2(ca+a+1)}$

Thus, $\dfrac{1}{a^2+2b^2+3}+\dfrac{1}{b^2+2c^2+3}+\dfrac{1}{c^2+2a^2+3}\le\dfrac{1}{2(ab+b+1)}+\dfrac{1}{2(bc+c+1)}+\dfrac{1}{2(ca+a+1)}$

I know this is really close to the answer but something messed with my brain and I cannot think any further. How do I continue and is there any better solution than this?

Answer 1

You've done all the hard work! Now you only need to prove $$\frac{1}2\left(\frac{1}{ab+b+1}+\frac{1}{bc+c+1}+\frac{1}{ca+a+1}\right)\leq\frac{1}2$$

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HINT: utilize $abc=1$. $$\frac{1}{ab+b+1}=\frac{c}{abc+bc+c}=\frac{c}{1+bc+c}$$

$$\frac{1}{ca+a+1}=\frac{bc}{abc^2+abc+bc}=\frac{bc}{c+1+bc}$$

Therefore, $$\frac{1}{ab+b+1}+\frac{1}{bc+c+1}+\frac{1}{ca+a+1}=?$$