Ahiles wrote: Let $ a,b,c$ be positive reals with $ ab + bc + ca = 3$. Prove that: \[ \frac {1}{1 + a^2(b + c)} + \frac {1}{1 + b^2(a + c)} + \frac {1}{1 + c^2(b + a)}\le \frac {1}{abc} \] It just mathlinks contest but weaker .

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Yes it is very similar and here there exist and easy solution . We have $ \sum\frac{1}{1+a^2(b+c)}\leq\sum\frac{1}{abc+a^2(b+c)}=\sum\frac{1}{3a}=\frac{1}{abc}$

silouan wrote: Yes it is very similar and here there exist and easy solution . We have $ \sum\frac {1}{1 + a^2(b + c)}\leq\sum\frac {1}{abc + a^2(b + c)} = \sum\frac {1}{3a} = \frac {1}{abc}$ Nice solution,silouan !

akai wrote: Nice solution,silouan ! Thank you akai .

Well the nice solution has already been posted... so I might as well note that there is a lower bound with Jensen's: Let $ f(x)=\frac{1}{x+k}$ which is convex for arbitrary $ k$. In particular set $ k=1-abc$. \[ \sum\frac{1}{1+a^2(b+c)}=\sum f(3a)\geq\frac{1}{a+b+c+\frac13-\frac{abc}{3}}\]

silouan wrote: Yes it is very similar and here there exist and easy solution . We have $ \sum\frac{1}{1+a^2(b+c)}\leq\sum\frac{1}{abc+a^2(b+c)}=\sum\frac{1}{3a}=\frac{1}{abc}$ how did you write first inequality? i didn't understand why abc <= 1? (abc small or equal = 1) sorry for my english.

Use am_gm $\sum_{cyc} ab \geq3 \sqrt[3] {a^2b^2c^2}$

From ma>=mg we have (ab+bc+ac)/3>=∛(a*a*b*b*c*c) <=> 3/3=1>=abc (ab+bc+ca=3) a*b*c<=1; We shrink the denominator replaces 1 with a*b*c. ∑1/(1+a*a(b+c) )<=∑1/(a*b*c+a*a(b+c) )=∑1/(a(ab+ac+bc))= 1/3 ∑ 1/a=1/3(1/a+1/b+1/c)= 1/3 (ab+bc+ac)/abc=1/3*3/abc=1/(a*b*c)

Ahiles wrote: Let $ a,b,c$ be positive reals with $ ab + bc + ca = 3$. Prove that: \[ \frac {1}{1 + a^2(b + c)} + \frac {1}{1 + b^2(a + c)} + \frac {1}{1 + c^2(b + a)}\le \frac {1}{abc}. \] \[ \frac {1}{1 + a^2(b + c)} + \frac {1}{1 + b^2(a + c)} + \frac {1}{1 + c^2(b + a)}\le Q \le \frac {1}{abc}. \] $Q=\frac{1}{3}(\frac{c(1+a)}{1+b}+\frac{a(1+b)}{c+1}+\frac{b(c+1)}{1+a})$

We have \[\sum_{cyc} \frac{1}{1+a^2(b+c)}\leq \frac{1}{abc} \iff \sum_{cyc}\frac{1}{3a}-\frac{1}{1+a(3-bc)}\geq 0 \iff \sum_{cyc} \frac{1-abc}{3a(1+3a-abc)} \geq 0\iff 1\geq abc\]which follows from \[1=\frac{ab+bc+ca}{3}\geq \sqrt[3]{a^2b^2c^2} \implies 1\geq abc.\]