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

AI Thread Summary
The discussion focuses on proving the inequality $$\frac{ab}{a+b+ab}+\frac{bc}{b+c+bc}+\frac{ca}{c+a+ca}\le \frac{a^2+b^2+c^2+6}{9}$$ for all positive real numbers a, b, and c. Participants share their approaches and insights, with one member praising another's method. The emphasis is on finding a valid proof that adheres to the conditions of the inequality. The conversation highlights the collaborative nature of problem-solving in mathematical discussions. Overall, the thread showcases a constructive exchange aimed at tackling a specific mathematical challenge.
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Prove $$\frac{ab}{a+b+ab}+\frac{bc}{b+c+bc}+\frac{ca}{c+a+ca}\le \frac{a^2+b^2+c^2+6}{9}$$ for all positive real $a,\,b$ and $c$ and $a,\,b,\,c\ne 0$.
 
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anemone said:
Prove $$\frac{ab}{a+b+ab}+\frac{bc}{b+c+bc}+\frac{ca}{c+a+ca}\le \frac{a^2+b^2+c^2+6}{9}$$ for all positive real $a,\,b$ and $c$ and $a,\,b,\,c\ne 0$.

Hint:

Note that $$\frac{ab}{a+b+ab}=\frac{1}{\frac{1}{a}+\frac{1}{b}+1}$$.
 
my solution:
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 $AP\geq GP$)
 
Well done Albert!(Cool) That is how I approached the problem as well!
 
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