<s>[INST] nice classical nonhomogeneous inequality Let $a,b,c$ be positive reals and $abc=1$. Prove that $$10(a^4+b^4+c^4)+21\ge 17(a^3+b^3+c^3).$$
I have found a solution using MV and I'm wondering if there is a nice solution.
 [/INST] This is just a solution, not necessarily a nice one. I started by simple rearrangement, and I got
$$ 10(a^4+b^4+c^4 - 3abc) \geq 17(a^3+b^3+c^3 - 3abc)$$ but I couldn't think of any well-known inequality that can prove this. So I decided to do straight calculus: 
Let $g(a,b,c) = 1-abc$ and $f(a,b,c)=10*(a^4+b^4+c^4)-17(a^3+b^3+c^3)$. By Lagrange Theorem, to find the extremal points we need to 
$$
  \begin{cases}
    g = 0\\ 
    \nabla f = \lambda \nabla g
  \end{cases}   \Leftrightarrow \begin{cases}
    abc = 1\\ 
    40a^3 - 51a^2 = \lambda bc\\ 
    40b^3 - 51b^2 = \lambda ac\\ 
    40c^3 - 51c^2 = \lambda ab
  \end{cases}
$$
which can be simplified to
$$
  \begin{cases}
    40a^4 - 51a^3 = \lambda \\ 
    40b^4 - 51b^3 = \lambda \\ 
    40c^4 - 51c^3 = \lambda 
  \end{cases}
$$
Subtracting first two equations in the system we get $40a^4-51a^3 = 40b^4-51b^3$. If $a=b=c=1$ we observe that $f(1,1,1)+21=0 \geq 0$. If $a=b\not =c$, then we may rewrite our system as
$$
  \begin{cases}
    a^2 c = 1\\
    40a^3 - 51a^2 - \lambda ac = 0\\ 
    40c^3 - 51c^2 - \lambda a^2 = 0
  \end{cases}
$$
After simple manipulations and eliminating $\lambda$, we get $40a^8+51a^7+51a^2-40=0$ which has two real roots $-1$ and $\approx 0.7527$ (and 6 other complex roots). The only one satisfying all of our conditions is $a=0.7527$. 
Setting the values back into our function yields $f\left(0.7527,0.7527,1.765\right) + 21 = 16.4973 \gt 0$. Now we only need to check the boundary case $f(0,0,1)+21 = 14 \gt >0$. By the Lagrange Theorem the minimum value of the function $f(a,b,c)$ with the bounding condition $g(a,b,c)$ is $-4.5027$.
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