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 $a+b+c=1$.

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!  
