prove this inequality by $abc=1$ Let $a,b,c>0$ and $abc=1$,show that
$$(a^{10}+b^{10}+c^{10})^2\ge 3(a^{14}+b^{14}+c^{14})$$
since
$$LHS=\sum \left(a^{20}+\dfrac{2}{a^{10}}\right)$$
it is prove
$$\sum_{cyc}\left(a^{20}+\dfrac{2}{a^{10}}-3a^{14}\right)\ge 0$$ not easy,and I found $x^{20}+2x^{-10}-3x^{14}$http://www.wolframalpha.com/input/?i=x%5E(20)%2B2x%5E(-10)-3x%5E(14)


We start with two identities that we can call Lamé-type identites:
$$(x+y+z)^5 - (x^5+y^5+z^5) = 5(x+y)(x+z)(y+z)(x^2+y^2+z^2+xy+xz+yz)=P$$
$$(x+y+z)^7 - (x^7+y^7+z^7) = 7(x+y)(x+z)(y+z)((x^2+y^2+z^2+xy+xz+yz)^2+xyz(x+y+z))=Q$$
Furthermore we have :
$$(a+b)(b+c)(c+a)=\frac{(a+b+c)^3-a^3-b^3-c^3}{3}$$
Or 
With the identity of Gauss we have :
$$a^3+b^3+c^3-3abc=(a+b+c)(a^2+b^2+c^2-ab-bc-ca)$$
So we deduce this :
$$(x+y+z)^5 - (x^5+y^5+z^5) = 5(\frac{(x+y+z)^3+(x+y+z)(xz+yz+yz-x^2-y^2-z^2)-3}{3})(x^2+y^2+z^2+xy+xz+yz)$$
And 
$$(x+y+z)^7 - (x^7+y^7+z^7)=7(\frac{(x+y+z)^3+(x+y+z)(xz+yz+yz-x^2-y^2-z^2)-3}{3})((x^2+y^2+z^2+xy+xz+yz)^2+xyz(x+y+z))$$
your inequality is equivalent to with ($xyz=1$) :
$$(x^5+y^5+z^5)^2\ge 3(x^7+y^7+z^7)$$
Or :
$$(x^5+y^5+z^5-(x+y+z)^5+(x+y+z)^5)^2\ge 3(x^7+y^7+z^7-(x+y+z)^7+(x+y+z)^7)$$
Or :
$$(-P+(x+y+z)^5)^2\geq 3(-Q+(x+y+z)^7)$$
Or :
$$-P\geq (3(-Q+(x+y+z)^7))^{0.5}-(x+y+z)^5$$
But :
$$-P=-(x+y+z)^5 + (x^5+y^5+z^5) = 5(\frac{-(x+y+z)^3+(x+y+z)(-xz-yz-yz+x^2+y^2+z^2)+3}{3})((x+y+z)^2-(xy+xz+yz))$$
$$-Q=-(x+y+z)^7 + (x^7+y^7+z^7)=7(\frac{(x+y+z)^3+(x+y+z)(xz+yz+yz-x^2-y^2-z^2)-3}{3})(-(-(x+y+z)^2+xy+zx+yz)^2-xyz(x+y+z))$$
Now we remark that if we put $x+y+z=\alpha$$\quad$ so $xy+yz+zx$ is maximal for $x=y=z=\frac{\alpha}{3}$
Furthermore in $-Q$ all the $xy+yz+zx$ are positive and in $-P$ there are negative so we can minimizing and maximazing the LHS and the RHS respectively
So we can reduce the problem to a one variable problem wich is easily solvable if we remark that :
$$xyz=1 \implies x+y+z\geq 3$$
