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.
 A: 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 $-4.5027$.
A: Here's another idea/solution!
Define the following function:
$$F(a,b,c) = 10(a^4 + b^4 + c^4) + 21 - 17(a^3 + b^3 + c^3)$$
First we will prove that:
$$F(a,b,c) - F(a,\sqrt{bc},\sqrt{bc}) \ge 0 \quad \quad \text{where $bc\ge 1$}$$
After substitution and cancelation it comes down to proving the first part of the solution. So I'll skip that.
Now it comes down to proving that:
$$F(a,t,t) \ge 0 \quad \quad \text{where $at^2 = 1$}$$
which is basically can be reduced to an one-variable polynomial and can be easily solved. We get:
$$F(a,t,t) = 20t^4 + 10a^4 + 21 - 17a^3 - 34t^3$$
Now substitute: $a=\frac 1{t^2}$ form the condition and we have:
$$F(a,t,t) = 20t^{12} - 34t^{31} - 21t^8 - 17t^2 + 10 = 20(t-1)^{12} + 206(t-1)^{11} + 946(t-1)^{10} + 2530(t-1)^9 + 4311(t-1)^8 + 4788(t-1)^7 + 3360(t-1)^6 + 1308(t-1)^5 + 150(t-1)^4 - 34(t-1)^3  + 20(t-1)^2 \ge 0$$
Obviously true since $t\ge 1$ this is true. Actually this is true regardless whether $t\ge 1$ or not, but I can't find a way to prove it explicitly. Also the big expression is nothing more than just normal polynomial division, but you have to be careful with all those numbers and it takes quite a bit of time.
Obviously we have inequality when: $a=b=c=1$
A: Let $a+b+c=3u$, $ab+ac+bc=3v^2$ and $abc=w^3$.
Hence, we need to prove that
$$10(81u^4-108u^2v^2+18v^4+12uw^3)+21w^4-17(27u^3-27uv^2+3w^3)\geq0$$ 
or $f(u)\geq0$, where
$$f(u)=270u^4-360u^2v^2+60v^4+40uw^3-153u^3w+153uv^2w-10w^4.$$
But by Schur $w^3\geq4uv^2-3u^3$.
Thus,
$$f'(u)=1080u^3-720uv^2+40w^3-459u^2w+153v^2w\geq$$
$$\geq621u^3-720uv^2+99w^3=9(69u^3-80uv^2+11w^3)\geq$$
$$\geq9(69u^3-80uv^2+44uv^2-33u^3)=324u(u^2-v^2)\geq0,$$
which says that $f$ is an increasing function.
Thus, it's enough to prove our inequality for a minimal value of $u$, 
which happens for equality case of two variables.
Let $b=a$ and $c=\frac{1}{a^2}$.
Id est, we need to prove that:
$$(a-1)^2(20a^{10}+6a^9-8a^8-22a^7-15a^6-8a^5-a^4+6a^3+13a^2+20a+1)\geq0,$$
which is true.
Done!
