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Yes, its true. W.L.O.G. $a\geq b \geq c$. Let $a+b+c=s,\ abc=p$ then $b=\left((s-a)+d\right)/2, c=\left((s-a)-d\right)/2$ with $d=b-c=\sqrt{(s-a)^2-4p/a}$. If we consider $b,c$ as a variable in $a,s,p$ we get $$ \begin{eqnarray} \frac{\partial (a^n+b^n+c^n)}{\partial a} &=& na^{n-1}+nb^{n-1}\frac{1}{2}\left(-1+\frac{\partial d}{\partial a}\right)+nc^{n-1}\frac{1}{2}\left(-1-\frac{\partial d}{\partial a}\right)\\ &=& \frac{n}{2}\left(\left(2a^{n-1}-b^{n-1}-c^{n-1}\right)+\frac{\partial d}{\partial a}\left(b^{n-1}-c^{n-1}\right) \right). \end{eqnarray} $$ Hence it is enough to prove that $$ \frac{\partial d}{\partial a}+\frac{2a^{n-1}-b^{n-1}-c^{n-1}}{b^{n-1}-c^{n-1}}\geq 0 $$ We have $$ \frac{\partial d}{\partial a}=\frac{a-s+2p/a^2}{d}=\frac{-b-c+2bc/a}{b-c}=\frac{2bc(b^{n-2}+b^{n-3}c+\dots)/a-(b+c)(b^{n-2}+b^{n-3}c+\dots)}{b^{n-1}-c^{n-1}} $$ Hence $$ \frac{\partial d}{\partial a}+\frac{2a^{n-1}-b^{n-1}-c^{n-1}}{b^{n-1}-c^{n-1}} =\frac{2bc(b^{n-2}+b^{n-3}c+\dots)/a+2a^{n-1}-(b+c)(b^{n-2}+b^{n-3}c+\dots)-(b^{n-1}+c^{n-1})}{b^{n-1}-c^{n-1}} $$ Taking the derivative by $a$ yields $(2(n-1)a^{n-2}-2bc/a^2(b^{n-2}+b^{n-3}c+\dots))/(b^{n-1}-c^{n-1})$ which is larger or equal to zero. Hence the expression above is minimal if $a$ is chosen minimal, i.e. $a=b$. In this case we have $\frac{\partial d}{\partial a}=-1=-\frac{2a^{n-1}-b^{n-1}-c^{n-1}}{b^{n-1}-c^{n-1}}$.