it is not really tough. You only need to use $\frac{1}{3a+4b+5c} \leq \frac{1}{36}\left(\frac{1}{a+b}+\frac{2}{a+c}+\frac{3}{b+c}\right)$. If I'm not wrong, the rest is easy.

I used expansion . It's equivalent to $\sum (60a^4+11a^3b-432a^2b^2+73ab^3+288a^2bc)\geq 0$ . I have an ungly proof for the last..... Do you have a better solution for the last ??

what is "the last"?

This one $\sum (60a^4+11a^3b-432a^2b^2+73ab^3+288a^2bc)\geq 0$ . BTW , Soarer your solution is very nice

hi, here is my solution : $x = 3a+4b+5c , y = 3b+4c+5a , z = 3c+4a+5b$ $x\geq12\sqrt[12]{a^3b^4c^5} \dots$ $\frac{1}{12}\times({\frac{ab}{\sqrt[12]{a^3b^4c^5}}+\dots)\geq S}$ $rearangement \Rightarrow \frac{a+b+c}{12}\geq S$ .

Soarer wrote: it is not really tough. You only need to use $\frac{1}{3a+4b+5c} \leq \frac{1}{36}\left(\frac{1}{a+b}+\frac{2}{a+c}+\frac{3}{b+c}\right)$. If I'm not wrong, the rest is easy. Great! Very elegant! to toetoe: you need to proove S<=... not S>=...

i have just edited the reply.

Soarer did you use some type of fudging to get to that inequality you posted. if so could fill in some details into how you used it. im just learning this nice techique. thanks

Recall the inequality $\sum \frac{ab}{a+b+2c} \leq \frac{3}{4}$ My idea comes from here, so I try to break $3a+4b+5c$ into $a+b$, $b+c$, $c+a$. Then you can let $3a+4b+5c = m(a+b)+n(b+c)+k(c+a)$ and find $m=1, n=3, c=2$

Soarer wrote: it is not really tough. You only need to use $\frac{1}{3a+4b+5c} \leq \frac{1}{36}\left(\frac{1}{a+b}+\frac{2}{a+c}+\frac{3}{b+c}\right)$. If I'm not wrong, the rest is easy. I have difficult in last step. I have obtained $\sum {a} \geq \sum \frac{ab}{a+b} + \sum \frac{ab}{c+b}$ which I cannot solve. Please a hint. Thank you very much.

The idea first comes from \[ \frac1x+\frac1y\geq\frac2{\sqrt{xy}}\geq\frac4{x+y}. \] This is used to solve \[ \frac{ab}{2c+a+b}+\frac{bc}{2a+b+c}+\frac{ca}{2b+c+a}\leq\frac{a+b+c}4. \] Following Soarer, and use $1/x+1/y\geq4/(x+y$ again, you get it Manlio.

Sorry, but if I use your idea I get $abc \sum (a) \geq \sum (ab)^2$ which is wrong. I cannot find my error.

As you said Manlio the inequality is equivalent to $\frac{ca+ab}{a+b}+\frac{bc+ab}{b+c}+\frac{bc+ca}{c+a}\leq a+b+c$ . Expand erything and you will have to prove that $b^3a+c^3b+a^3c\geq abc(a+b+c)$ Divide with $abc$ and you will have to prove that $\frac{b^2}{c}+\frac{c^2}{a}+\frac{a^2}{b}\geq a+b+c$ which is true by Cauchy

manlio wrote: Soarer wrote: it is not really tough. You only need to use $\frac{1}{3a+4b+5c} \leq \frac{1}{36}\left(\frac{1}{a+b}+\frac{2}{a+c}+\frac{3}{b+c}\right)$. If I'm not wrong, the rest is easy. I have difficult in last step. I have obtained $\sum {a} \geq \sum \frac{ab}{a+b} + \sum \frac{ab}{c+b}$ which I cannot solve. Please a hint. Thank you very much. $\sum \frac{ab}{a+b} \leq \sum \frac{ac}{c+b}$ by rearrangement. Then we just add up.

Thank you very much

It's teribble if someone can give solution with SOS.

Lovasz wrote: It's teribble if someone can give solution with SOS. I solved this problem with SOS . The basic idea is this expration \[ \frac{ab}{3a+4b+5c}-\frac{ab(5a+4b+3c)}{16(a+b+c)^2}=\frac{ab(a-c)^2}{16(a+b+c)^2(3a+4b+5c)} \]

Prove that if $a,b,c,d>0,$ then \[ \frac{abc}{3a+4b+5c+6d}+\frac{bcd}{3b+4c+5d+6a}+\frac{cda}{3c+4d+5a+6b}+\frac{dab}{3d+4a+5b+6c}\]\[\le \frac{a^2+b^2+c^2+d^2}{18}. \]

My solution: multiply everything by $P= \prod(3a+4b+5c) $ and you get $ (a+b+c)P >= \sum abP/(3a+4b+5c) $ and than you use Muirheads inequality and it's done . Equality for $ a=b=c$.

Soarer wrote: it is not really tough. You only need to use $\frac{1}{3a+4b+5c} \leq \frac{1}{36}\left(\frac{1}{a+b}+\frac{2}{a+c}+\frac{3}{b+c}\right)$. If I'm not wrong, the rest is easy. How do you finish the proof from this?

Duarti wrote: How do you finish the proof from this?

Sorry about this

Corrected proof: \[\frac{1}{36}\sum ab\left(\frac{1}{a+b}+\frac{2}{a+c}+\frac{3}{b+c}\right) = \frac{1}{36}\left(\sum\left(\frac{a(b+c)}{a+b}\right) + 2a+2b+2c \right) \stackrel{\text{Rearrangement}}{\leq} \frac{1}{36}(3a+3b+3c) = \frac{a+b+c}{12}\] Ok, the summation is very confusing, I'll check it later, the above might still be incorrect.

Vrangr, It should be reversed. No?

imortal wrote: Prove that if $a,b,c>0,$ then \[ \frac{ab}{3a+4b+5c}+\frac{bc}{3b+4c+5a}+\frac{ca}{3c+4a+5b}\le \frac{a+b+c}{12}. \] Nikolai Nikolov two lines down

Observe by Titu that \[\frac1{3a+4b+5c}=\frac1{(a+b)+2(a+c)+3(b+c)} \le\frac1{36}\left(\frac1{a+b}+\frac2{a+c}+\frac3{b+c}\right),\]hence by rearrangement we have \begin{align*} \sum_\mathrm{cyc}\frac{ab}{3a+4b+5c} &\le\sum_\mathrm{cyc}\frac{ab}{36}\left(\frac1{a+b}+\frac2{a+c}+\frac3{b+c}\right)\\ &=\frac1{36}\sum_\mathrm{cyc}\frac{ab+2bc+3ca}{a+b}\\ &=\frac1{36}\sum_\mathrm{cyc}\left(3c+\frac{ab}{a+b}-\frac{ac}{a+b}\right) \le\frac{a+b+c}{12}. \end{align*}

Here's some motivation behind the above solution. It seems pretty natural to conjecture that if $3,4,5$ are replaced by any $x,y,z \in \mathbb R^+$, then the inequality still holds true. In particular, the conjecture is $$\frac{ab}{xa + yb + zc} + \frac{bc}{xb + yc + za} + \frac{ca}{xc + ya + zb} \le \frac{a+b+c}{x+y+z}$$But if we fix $x,y$ and tend $z \to \infty$, then our conjecture implies $$\frac{ab}{c} + \frac{bc}{a} + \frac{c}{ab} \le a+b+c$$But this clearly isn't true (in fact, the inequality is not true if the sign is reversed). So our conjecture is false. So as seen above, if $5$ was replaced by a larger number, then the inequality won't be true. So we somehow have to exploit the fact that $5$ is near to $3,4$. This gives the motivation to pair $5c$ with $3a,4b$.

toetoe wrote: hi, here is my solution : $x = 3a+4b+5c , y = 3b+4c+5a , z = 3c+4a+5b$ $x\geq12\sqrt[12]{a^3b^4c^5} \dots$ $\frac{1}{12}\times({\frac{ab}{\sqrt[12]{a^3b^4c^5}}+\dots)\geq S}$ $rearangement \Rightarrow \frac{a+b+c}{12}\geq S$ . I don’t think that is correct since $\sum_{cyc} \frac{ab}{\sqrt[12]{a^3b^4c^5}} \le \frac{a + b + c}{12}$ is visibly not true as the RHS is more “mixed” than the LHS. Does anyone have a solution besides @Soarer’s and @TheUltimate123’s very beautiful intercalation? It’s a bit difficult to see.