<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] Another way.
By C-S and P-M we obtain:
$$\sum_{cyc}\frac{a^2}{a+2b^2}-1=\sum_{cyc}\frac{a^4}{a^3+2a^2b^2}-1\geq\frac{(a^2+b^2+c^2)^2}{\sum\limits_{cyc}(a^3+2a^2b^2)}-1=$$
$$=\frac{\sum\limits_{cyc}(a^4-a^3)}{\sum\limits_{cyc}(a^3+2a^2b^2)}=\frac{9(a^4+b^4+c^4)-(a^3+b^3+c^3)(\sqrt{a}+\sqrt{b}+\sqrt{c})^2}{9\sum\limits_{cyc}(a^3+2a^2b^2)}\geq0.$$
Because by P-M $$\sqrt[4]{\frac{a^4+b^4+c^4}{3}}\geq\sqrt[3]{\frac{a^3+b^3+c^3}{3}}$$ or $$\sqrt[4]{\left(\frac{a^4+b^4+c^4}{3}\right)^3}\geq\frac{a^3+b^3+c^3}{3}$$ or $$\sqrt[4]{3(a^4+b^4+c^4)^3}\geq a^3+b^3+c^3$$ and
$$\sqrt[4]{\frac{a^4+b^4+c^4}{3}}\geq\left(\frac{\sqrt{a}+\sqrt{b}+\sqrt{c}}{3}\right)^2$$ or
$$\sqrt[4]{2187(a^4+b^4+c^4)}\geq(\sqrt a+\sqrt b+\sqrt c)^2.$$
Id est, $$9(a^4+b^4+c^4)=\sqrt[4]{3(a^4+b^4+c^4)^3}\cdot\sqrt[4]{2187(a^4+b^4+c^4)}\geq$$
$$\geq(a^3+b^3+c^3)(\sqrt{a}+\sqrt{b}+\sqrt{c})^2.$$
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