Let $a,b,c\geq 0$ with $ a+b+c=3$ ,show that

\[ \sqrt{a^2+bc+2}+\sqrt{b^2+ca+2}+\sqrt{c^2+ab+2}\geq 6 \]

(lhclp008)

Proof

by Holder inequality
\[ \left(\sum{\sqrt{a^2+bc+2}}\right)^2\left[\sum{\frac{(a^2+2bc+9)^3}{a^2+bc+2}}\right]\geq [(a+b+c)^2+27]^3=36^3 \]
Therefore,it's suffice to check
\[  \frac{(a^2+2bc+9)^3}{a^2+bc+2}+\frac{(b^2+2ca+9)^3}{b^2+ca+2}+\frac{(c^2+2ab+9)^3}{c^2+ab+2}\leq 1296 \]

Which can be checked by Muirhead's theorem.

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also,we have

\[ \sqrt{3a^2+4bc+9}+\sqrt{3b^2+4ca+9}+\sqrt{3c^2+4ab+9}\geq 12. \]

in the same conditions.

(quyhktn-qa1)

http://www.artofproblemsolving.com/Forum/viewtopic.php?f=52&t=538752

it's suffices to prove

$$ \frac{(a^2+2bc+9)^3}{3a^2+4bc+9}+\frac{(b^2+2ca+9)^3}{3b^2+4ca+9}+\frac{(c^2+2ab+9)^3}{3c^2+4ab+9}\le 324 $$

Which can also be checked by Muirhead's theorem.