<s>[INST] Prove $\sqrt{a+b}+\sqrt{b+c}+\sqrt{c+a}\ge \sqrt{7\left(a^3+b^3+c^3+a^2b+b^2c+c^2a+b^2a+a^2c+c^2b\right)-3}$ with $a^2+b^2+c^2=1$ and $a,b,c\ge 0$ With $a,b,c\ge 0$ and $a^2+b^2+c^2=1$, prove $\sqrt{a+b}+\sqrt{b+c}+\sqrt{c+a}\ge \sqrt{7\left(a^3+b^3+c^3+a^2b+b^2c+c^2a+b^2a+a^2c+c^2b\right)-3}$.
I don't know when equality holds for this problem, can you give me some hint? Thank!
 [/INST] It suffices to prove that
$$(\sqrt{a + b} + \sqrt{b + c} + \sqrt{c + a})^2 \ge 7(a + b + c) - 3$$
or
$$2a + 2b + 2c + \sum_{\mathrm{cyc}} 2\sqrt{(a + b)(b + c)} \ge 7(a + b + c) - 3.$$
Using Cauchy-Bunyakovsky-Schwarz inequality and GM-HM inequality, we have
$$\sqrt{(a + b)(b + c)} \ge b + \sqrt{ca} \ge b + \frac{2ca}{c + a}
\ge b + \frac{2ca}{a + b + c}.$$
It suffices to prove that
$$4a + 4b + 4c + \frac{4(ab + bc + ca)}{a + b + c} \ge 7(a + b + c) - 3$$
or
$$4a + 4b + 4c + \frac{2 \cdot [(a + b + c)^2 - 1]}{a + b + c} \ge 7(a + b + c) - 3$$
or
$$(a + b + c - 1)(2 - a - b - c) \ge 0$$
which is true (easy).
We are done.
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