<s>[INST] Prove the inequality $\frac a{a+b^2+c^2} + \frac b{b+c^2+a^2}+\frac c{c+a^2+b^2} \le 1$ $a, b, c > 0$ such that $abc=1$, prove inequality (Lagrang not allowed here because I am still in grade $9^{th}$):
$$P=\frac a{a+b^2+c^2} + \frac b{b+c^2+a^2}+\frac c{c+a^2+b^2} \le 1$$ 
I have tried to use AM-GM, that I have:
$(a+b^2+c^2)(a+1+1)\ge (a+b+c)^2$, so $$P \le \frac{a(a+2)+b(b+2)+c(c+2)}{(a+b+c)^2}$$ but if $$\frac{a(a+2)+b(b+2)+c(c+2)}{(a+b+c)^2} \le 1 \Rightarrow ab + bc + ca \ge a+b+c \ (\text{not true})$$
P/S: Can someone tell me what good inequalities books that I should buy ?
 [/INST] Let $a = \frac{x^2}{yz}$ and so forth. Expanding (*):
$$(x^4 yz + y^6 + z^6)(y^4 xz + x^6 + z^6)(z^4 xy + x^6 + y^6) - x^4yz (y^4 xz + x^6 + z^6)(z^4 xy + x^6 + y^6) - y^4xz (x^4yz + y^6 + z^6)(z^4 xy + x^6 + y^6) - z^4xy(x^4yz + y^6 + z^6)(y^4 xz + x^6 + z^6) $$
at quickmath.com gives (**)
$$x^{12}y^6 + x^{12}z^6 + y^{12}x^6 + y^{12}z^6 + z^{12}x^6 + z^{12}y^6 - x^2y^5z^{11} - x^2y^{11}z^5 - y^2x^5z^{11} - y^2x^{11}z^5 - z^2x^5y^{11} - z^2x^{11}y^5$$
The given inequality is equivalent to $(*) \ge 0$, same as $(**) \ge 0$, which follows from Muirhead inequality and $[12,6,0] \succ [11,5,2]$.

An aesthetically more appealing solution is note that 
$$\frac{a}{a+b^2+c^2} \leq \frac{a^r}{a^r+b^r+c^r}$$
for $r = 4/3$. To show this, note that after clearing denominator we need to show 
$$\begin{eqnarray} a(b^r+c^r) \leq a^r(b^2+c^2)\\
\Leftrightarrow  (bc)^{r-1}(b^r+c^r) = a^{1-r}(b^r+c^r) \leq b^2+c^2 \end{eqnarray}$$
Now $(bc)^{1/3}(b^{4/3} + c^{4/3}) \leq b^2 + c^2$ follows from AM-GM and power mean inequality.
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