Prove the inequality $\frac a{a+b^2}+\frac b{b+c^2}+\frac c{c+a^2}\le\frac14\left(\frac1a+\frac1b+\frac1c\right)$ Let $a,b,c>0; a+b+c=1$. Prove the inequality
$$\frac a{a+b^2}+\frac b{b+c^2}+\frac c{c+a^2}\le\frac14\left(\frac1a+\frac1b+\frac1c\right)$$
My work so far:
I tried AM-GM and used fact $a+b+c=1$.
 A: An homogenization and a full expanding give:
$$\sum\limits_{cyc}(a^6b^2+2a^5b^3+a^5c^3+2a^4b^4+a^6bc-2a^5b^2c+4a^5c^2b+$$
$$+a^4b^3c+4a^4c^3b-7a^4b^2c^2-7a^3b^3c^2)\geq0$$
which is obvious by AM-GM.
A: Also we can us C-S and $uvw$:
We need to prove that $$\sum\limits_{cyc}\left(\frac{a}{a+b^2}-1\right)\leq\frac{ab+ac+bc}{4abc}-\frac{3}{a+b+c}$$ or
$$\sum\limits_{cyc}\frac{b^2}{a+b^2}+\frac{ab+ac+bc}{4abc}-\frac{3}{a+b+c}\geq0$$
Now by C-S:
$$\sum\limits_{cyc}\frac{b^2}{a+b^2}\geq\frac{(a+b+c)^2}{a+b+c+a^2+b^2+c^2}=\frac{a+b+c}{(a+b+c)^2+a^2+b^2+c^2}$$
Id est, it remains to prove that
$$\frac{a+b+c}{(a+b+c)^2+a^2+b^2+c^2}+\frac{ab+ac+bc}{4abc}-\frac{3}{a+b+c}\geq0$$
Let $a+b+c=3u$, $ab+ac+bc=3v^2$ and $abc=w^3$.
Hence, we need to prove that $f(w^3)\geq0$, where $f$ is a decreasing function, 
which says that it's enough to prove $f(w^3)\geq0$ for a maximal value of $w^3$.
Since $a$, $b$ and $c$ are positive roots of the equation
$$(x-a)(x-b)(x-c)=0$$ or
$$w^3=x^3-3ux^2+3v^2x$$
we see that $w^3$ gets a maximal value for equality case of two variables
and since $f(w^3)\geq0$ is an homogeneous inequality, it's enough to prove it for $b=c=1$, which gives 
$$(a-1)^2(2a^2+3a+6)\geq0$$
Done!  
A: $$\frac{a}{a+b^2}= \frac{a}{a^2+ab+ac+b^2} = \frac{a}{(a^2+ac)+(ab+b^2)} \le \frac{a}{4} \cdot \left( \frac{1}{a^2+ac}+\frac{1}{ab+b^2} \right)=$$
$$= \frac{1}{4}\cdot \left( \frac{1}{a+c}+\frac{1}{b}-\frac{1}{a+b} \right)$$
$$\frac{a}{a+b^2}+\frac{b}{b+c^2}+\frac{c}{c+a^2} \le \frac{1}{4}\cdot \left( \frac{1}{a}+\frac{1}{b}+\frac{1}{c} \right)$$
