Prove inequality $ (a+b+c)(a^7+b^7+c^7)\ge(a^5+b^5+c^5)(a^3+b^3+c^3)$ for $a,b,c\ge 0$ Prove that:
$$ (a+b+c)(a^7+b^7+c^7)\ge(a^5+b^5+c^5)(a^3+b^3+c^3)$$
I already know that this can be proven using Cauchy Schwarz, but I don't really see how to apply it here. I'm looking for hints. 
 A: If all of $a$, $b$ and $c$ are zero, then the inequality holds trivially, and so we can assume that at least one of $a$, $b$, and $c$ are positive.
The Cauchy-Schwarz inequality gives us that for any natural number $n$,
$$
 \left(a^n + b^n + c^n\right)\left(a^{n+2} + b^{n+2} + c^{n+2} \right)
 \geq \left(a^{n+1} + b^{n+1} + c^{n+1} \right)^2
$$
and so
$$
  \frac{a^n + b^n + c^n}{a^{n+1} + b^{n+1} + c^{n+1}} \geq \frac{a^{n+1} + b^{n+1} + c^{n+1}}{a^{n+2} + b^{n+2} + c^{n+2}}.
$$
We thus have that
$$
  \frac{a + b + c}{a^2 + b^2 + c^2} \geq \frac{a^2 + b^2 + c^2}{a^3 + b^3 + c^3} \geq \frac{a^3 + b^3 + c^3}{a^4 + b^4 + c^4} \geq \frac{a^4 + b^4 + c^4}{a^5 + b^5 + c^5} \geq \frac{a^5 + b^5 + c^5}{a^6 + b^6 + c^6},
$$
and in a similar way, it follows that
$$
  \frac{a^2 + b^2 + c^2}{a^3 + b^3 + c^3} \geq \frac{a^6 + b^6 + c^6}{a^7 + b^7 + c^7}.
$$
We thus have that
$$
  \frac{a + b + c}{a^3 + b^3 + c^3} = \frac{a + b + c}{a^2 + b^2 + c^2} \cdot \frac{a^2 + b^2 + c^2}{a^3 + b^3 + c^3} \geq \frac{a^5 + b^5 + c^5}{a^6 + b^6 + c^6} \cdot \frac{a^6 + b^6 + c^6}{a^7 + b^7 + c^7} = \frac{a^5 + b^5 + c^5}{a^7 + b^7 + c^7},
$$
which implies that
$$
  \left(a + b + c\right) \left(a^7 + b^7 + c^7 \right) \geq \left(a^5 + b^5 + c^5\right) \left(a^3 + b^3 + c^3 \right),
$$
as required.
A: I do not see a direct application of Cauchy-Schwarz. Anyway after the multiplication, we obtain
$$\sum_{sym}a^7b\geq \sum_{sym}a^5b^3$$
which holds by Muirhead's inequality.
A: Using Holder's inequality, we have that
\begin{align*}
a^5+b^5+c^5 &= a^{\frac{1}{3}}a^{\frac{14}{3}} + b^{\frac{1}{3}}b^{\frac{14}{3}}+c^{\frac{1}{3}}c^{\frac{14}{3}}\\
&\le (a+b+c)^{\frac{1}{3}}(a^7+b^7+c^7)^{\frac{2}{3}},
\end{align*}
and
\begin{align*}
a^3+b^3+c^3 &= a^{\frac{2}{3}}a^{\frac{7}{3}} + b^{\frac{2}{3}}a^{\frac{7}{3}}+c^{\frac{2}{3}}a^{\frac{7}{3}}\\
&\le (a+b+c)^{\frac{2}{3}}(a^7+b^7+c^7)^{\frac{1}{3}}.
\end{align*}
Multiplying them together, we are done.
A: Expanding the two sides and cancelling like terms gives
$$a^7b+ab^7+b^7c+bc^7+c^7a+ca^7\geq a^5b^3+a^3b^5+b^5c^3+b^3c^5+c^5a^3+c^3a^5.$$
This follows from Muirhead's inequality, where we set $n=2, a_1=5,a_2=3,b_1=7$, and $b_2=1$.
