<s>[INST] Prove that if $\sqrt{a}+\sqrt{b}+\sqrt{c}=3$ then $ \frac{a^{2}}{a+2 b^{2}}+\frac{b^{2}}{b+2 c^{2}}+\frac{c^{2}}{c+2 a^{2}} \geq 1 $ Question -
Let $a, b, c$ be positive real numbers such that $\sqrt{a}+\sqrt{b}+\sqrt{c}=3$ . Prove that
$$
\frac{a^{2}}{a+2 b^{2}}+\frac{b^{2}}{b+2 c^{2}}+\frac{c^{2}}{c+2 a^{2}} \geq 1
$$
My try
$$
\frac{a^{2}}{a+2 b^{2}}=a-\frac{2 a b^{2}}{a+2 b^{2}} \geq a-\frac{2 a b^{2}}{3 \sqrt[3]{a b^{4}}}=a-\frac{2(a b)^{2 / 3}}{3}
$$
which implies that
$$
\sum_{c y c} \frac{a^{2}}{a+2 b^{2}} \geq \sum_{c y c} a-\frac{2}{3} \sum_{c y c}(a b)^{\frac{2}{3}}
$$
It suffices to prove that
$$
(a b)^{2 / 3}+( b c)^{2 / 3}+\left (c a)^{2 / 3} \leq 3\right.
$$
beacuse we can easily get that $\sum a \ge 3$
but i am not able to prove it..
note that we have to prove this using only am-gm or weighted am-gm or power mean or any kind of means inequality because author did not introduce any advance inequality yet...
any hints ???
thankyou
 [/INST] Elaborating on the above solution: 
We first wish to prove that the last inequality holds. Schur's inequality in its simplest form states that: For non-negative real numbers $x,y,z$, we have $x^3+y^3+z^3+3xyz \geq xy(x+y)+xz(x+z)+yz(y+z)$.
Now, expanding the R.H.S., we obtain: $(X+Y+Z)^3+9XYZ$ 
$=X^3+3X^2Y+3X^2Z+3XY^2+6XYZ+3XZ^2+Y^3+3Y^2Z+3YZ^2+Z^3+9XYZ$
$=X^3+Y^3+Z^3+3XYZ+12XYZ+3X^2Y+3X^2Z+3XY^2+3XZ^2+3Y^2Z+3YZ^2$
$\geq XY(X+Y) +XZ(X+Z) + YZ(Y+Z)+12XYZ+3X^2Y+3X^2Z+3XY^2+3XZ^2+3Y^2Z+3YZ^2$, 
$=X^2Y+XY^2+X^2Z+XZ^2+Y^2Z+YZ^2+12XYZ+3X^2Y+3X^2Z+3XY^2+3XZ^2+3Y^2Z+3YZ^2$
$=4X^2Y+4XY^2+4X^2Z+4XZ^2+4Y^2Z+4YZ^2+12XYZ$
$=4(X^2Y+XYZ+XY^2+X^2Z+XYZ+XZ^2+Y^2Z+XYZ+YZ^2)$ 
$=4[XY(X+Y+Z)+XZ(X+Y+Z)+YZ(X+Y+Z)]$
$=4(X+Y+Z)(XY+YZ+ZX)$
Simply flipping the inequality sign gives us: $4(X+Y+Z)(XY+YZ+ZX) \leq (X+Y+Z)^3+9XYZ$, so we are done.
Finally, we observe that this inequality implies the second last inequality. With $X+Y+Z=3$, we have:
$4(X+Y+Z)(XY+YZ+ZX) \leq (X+Y+Z)^3+9XYZ$
$\Rightarrow 4(XY+YZ+ZX) \leq  \ (X+Y+Z)^2+3XYZ$
$\Rightarrow 4(XY+YZ+ZX) -3XYZ  \leq  \ (X+Y+Z)^2 = 3^2 =9$ 
(Q.E.D.)
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