Inequality challenge for all positive (but not zero) real a, b and c

[anemone]

Prove \(\displaystyle \frac{ab}{a+b+ab}+\frac{bc}{b+c+bc}+\frac{ca}{c+a+ca}\le \frac{a^2+b^2+c^2+6}{9}\) for all positive real $a,\,b$ and $c$ and $a,\,b,\,c\ne 0$.

 

[anemone]

Prove \(\displaystyle \frac{ab}{a+b+ab}+\frac{bc}{b+c+bc}+\frac{ca}{c+a+ca}\le \frac{a^2+b^2+c^2+6}{9}\) for all positive real $a,\,b$ and $c$ and $a,\,b,\,c\ne 0$.

Hint:

Spoiler

Note that \(\displaystyle \frac{ab}{a+b+ab}=\frac{1}{\frac{1}{a}+\frac{1}{b}+1}\).

 

[Albert1]

my solution:

Spoiler

using $AP\geq HP$
We have :$\\
\dfrac{ab}{a+b+ab}\leq \dfrac{a+b+1}{9}---(1)\\
\dfrac{bc}{b+c+bc}\leq \dfrac{b+c+1}{9}---(2)\\
\dfrac{ca}{c+a+ca}\leq \dfrac{c+a+1}{9}---(3)\\
(1)+(2)+(3):\dfrac {ab}{a+b+ab}+\dfrac {bc}{b+c+bc}+\dfrac {ca}{c+a+ca}\leq\dfrac{2a+2b+2c+3}{9}\leq \dfrac{a^2+1+b^2+1+c^2+1+3}{9}= \dfrac{a^2+b^2+c^2+6}{9}$
(using $AP\geq GP$)

 

[anemone]

Well done Albert!(Cool) That is how I approached the problem as well!