Let $a,b,c>0$ and $n$ be postive integer,show that $$a\sqrt{b^2+(n^2-1)c^2}+b\sqrt{c^2+(n^2-1)a^2}+c\sqrt{a^2+(n^2-1)b^2}\le (a+b+c)^2+(n-3)(ab+bc+ac)$$
For $n=1$ it suffices to show that $$ab+bc+ac\le (a+b+c)^2-2(ab+bc+ac)=a^2+b^2+c^2$$ this is equivalent to the obvious inequality $$(a-b)^2+(b-c)^2+(c-a)^2\ge 0$$ But How to prove by $n\ge 2?$
For $n\geq2$ we have $\sqrt{a^2+(n^2-1)b^2}\leq\frac{a^2+(n-1)ab+(n^2-n)b^2}{a+(n-1)b}\Leftrightarrow b^2(a-b)^2\geq0$.
Thus, it remains to prove that $$\frac{c(a^2+(n-1)ab+(n^2-n)b^2)}{a+(n-1)b}\leq\sum\limits_{cyc}(a^2+(n-1)ab)$$ or $$n(n-2)^2\sum\limits_{cyc}(a^3bc-a^2b^2c)+(n-1)\sum\limits_{cyc}(a^4b+(n-1)a^4c+(n-3)a^3b^2-(2n-3)a^3c^2)\geq0$$ for which it's enough to prove that $$\sum\limits_{cyc}(a^4b-a^4c-3a^3b^2+3a^3c^2)+n\sum\limits_{cyc}(a^4c+a^3b^2-2a^3c^2)\geq0$$ and since $\sum\limits_{cyc}(a^4c+a^3b^2-2a^3c^2)=\sum\limits_{cyc}(b^4a-2b^3a^2+a^3b^2)=\sum\limits_{cyc}ab^2(a-b)^2\geq0$,
it remains to prove that $$\sum\limits_{cyc}(a^4b-a^4c-3a^3b^2+3a^3c^2)+2\sum\limits_{cyc}(a^4c+a^3b^2-2a^3c^2)\geq0$$ or $$\sum\limits_{cyc}(a^4b+a^4c-a^3b^2-a^3c^2)\geq0$$ which is Muirhead.
Done!