Exercise 5.4.1
\[\sum_{cyc}a\sqrt{2b+c^2}\le \sqrt{\sum_{cyc}a\sum_{cyc}(2ab+c^2a)}=\sqrt{2\sum_{cyc}a\sum_{cyc}ab+3\sum_{cyc}a^2b}=\sqrt{5\sum_{cyc}a^2b+2\sum_{cyc}ab^2+6ab c}\]
\[\le \sqrt{3\sum_{cyc}a^2b+\sum_{cyc}(a^3+ab^2)+2\sum_{cyc}ab^2+6abc}=\sqrt{(\sum_{cyc}a)^3}=3\sqrt{3}\]
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