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$a$, $b$ and $c$ are positives such that $3a + 4b + 5c = 12$. Calculate the maximum value of $$\frac{ab}{ab + a + b} + \frac{2ca}{ca + c + a} + \frac{3bc}{bc + b + c}$$

I want to know if there are any other solutions that are more practical. This came from my homework my teacher gave today. I have given my solution down below if you want to check out.

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We have that $$3a + 4b + 5c = 12$$

$$\implies (a + b) + 2(c + a) + 3(b + c) = 12 \implies \sqrt{ab} + 2\sqrt{ca} + 3\sqrt{bc} \le 6$$

In addition to that, $$\frac{ab}{ab + a + b} + \frac{2ca}{ca + c + a} + \frac{3bc}{bc + b + c} \le \frac{ab}{ab + 2\sqrt{ab}} + \frac{2ca}{ca + 2\sqrt{ca}} + \frac{3bc}{bc + 2\sqrt{bc}}$$

$$ = \frac{\sqrt{ab}}{\sqrt{ab} + 2} + \frac{2\sqrt{ca}}{\sqrt{ca} + 2} + \dfrac{3\sqrt{bc}}{\sqrt{bc} + 2} = 6 - 2\left(\frac{1}{\sqrt{ab} + 2} + \frac{2}{\sqrt{ca} + 2} + \frac{3}{\sqrt{bc} + 2}\right)$$

$$\le 6 - \frac{72}{\sqrt{ab} + 2\sqrt{ca} + 3\sqrt{bc} + 12} \le 6 - \frac{72}{6 + 12} = 2$$

The equality sign occurs when $a = b = c = 1$.

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  • $\begingroup$ Very nice solution! +1 $\endgroup$ – Michael Rozenberg Jun 6 at 18:05
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Also, we can use C-S and AM-GM: $$\frac{ab}{ab + a + b} + \frac{2ca}{ca + c + a} + \frac{3bc}{bc + b + c}\leq$$ $$\leq\frac{ab}{(1+2)^2}\left(\frac{1^2}{ab}+\frac{2^2}{a+b}\right)+\frac{2ca}{9}\left(\frac{1}{ca}+\frac{4}{c+a}\right)+\frac{3bc}{9}\left(\frac{1}{bc}+\frac{4}{b+c}\right)=$$ $$=\frac{2}{3}+\frac{4}{9}\cdot\frac{ab}{a+b}+\frac{8}{9}\cdot\frac{ca}{c+a}+\frac{4}{3}\cdot\frac{bc}{b+c}\leq$$

$$\leq\frac{2}{3}+\frac{4}{9}\cdot\frac{a+b}{4}+\frac{8}{9}\cdot\frac{c+a}{4}+\frac{4}{3}\cdot\frac{b+c}{4}=\frac{2}{3}+\frac{3a+4b+5c}{9}=2.$$

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