prove this inequality by $abc=1$ Let $a,b,c&gt;0$ 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 can use the Vasc's EV-Method, Corollary 1.8 (b):

https://www.emis.de/journals/JIPAM/images/059_06_JIPAM/059_06.pdf

Indeed, we need to prove that:
$$(x^5+y^5+z^5)^2\geq3xyz(x^7+y^7+z^7),$$ where $x$, $y$ and $z$ they are positives.

Now, let $x^5+y^5+z^5=constant$ and $x^7+y^7+z^7=constant.$

Thus, $xyz$ gets a maximal value for equality case of two variables 

and since the last inequality is homogeneous, it's enough to assume $y=z=1$, which gives
$$(x-1)^2(x^8+2x^7-2x^5-4x^4-2x^3+2x+4)\geq0$$ or
$$(x-1)^2((x^4-2)^2+2x(x^2-1)(x^4-1))\geq0.$$
Done!

Also, the BW helps (https://math.stackexchange.com/tags/buffalo-way/info), but it's a very complicated solution:

Let $x=\min\{x,y,z\}$, $y=x+u$ and $z=x+v$.

Hence, $$(x^5+y^5+z^5)^2-3xyz(x^7+y^7+z^7)=(u^2-uv+v^2)x^8-8(u^3-2u^2v-2uv^2+v^3)x^7+$$
$$+4(5u^4-17u^3v+50u^2v^2-17uv^3+5v^4)x^6+$$
$$+8(11u^5-20u^4v+25u^3v^2+25u^2v^3-20uv^4+11v^5)x^5+$$
$$+2(63u^6-79u^5v+50u^4v^2+100u^3v^3+50u^2v^4-79uv^5+63v^6)x^4+$$
$$+4(24u^7-21u^6v+5u^5v^2+25u^4v^3+25u^3v^4+5u^2v^5-21uv^6+24v^7)x^3+$$
$$+(42u^8-24u^7v+20u^5v^3+50u^4v^4+20u^3v^5-24uv^7+42v^8)x^2+$$
$$+(10u^9-3u^8v+10u^5v^4+10u^4v^5-3uv^8+10v^9)x+(u^5+v^5)^2\geq0$$
because for this it's enough to prove that $$4(u^3-2u^2v-2uv^2+v^3)^2-(u^2-uv+v^2)(5u^4-17u^3v+50u^2v^2-17uv^3+5v^4)\leq0,$$
which is true.

Indeed, let $u^2+v^2=2kuv$.

Thus, $k\geq1$ and we need to prove that
$$4(u^2+2uv+v^2)(u^2-3uv+v^2)^2\leq(u^2-uv+v^2)(5(u^4+v^4)-17uv(u^2+v^2)+50u^2v^2)$$ or
$$8(k+1)(2k-3)^2\leq(2k-1)(5(4k^2-2)-34k+50)$$ or
$$4k^3-12k^2+69k-56\geq0,$$ which is obvious for $k\geq1$.

Also, we can use SOS here.

Indeed, $$(x^5+y^5+z^5)^2-3xyz(x^7+y^7+z^7)=\sum_{cyc}(x^{10}+2x^5y^5-3x^8yz)=$$
$$=\sum_{cyc}\left(x^{10}+x^7y^3+x^7z^3-3x^8yz+2x^5y^5-x^7y^3-x^7z^3\right)=$$
$$=(x^7+y^7+z^7)(x^3+y^3+z^3-3xyz)-\sum_{cyc}x^3y^3(x^2-y^2)^2=$$
$$=\sum_{cyc}(x-y)^2\left(\frac{1}{2}(x+y+z)(x^7+y^7+z^7)-x^3y^3(x+y)^2\right).$$
Now, let $x\geq y\geq z$.

Thus, $$\frac{1}{2}(x+y+z)(x^7+y^7+z^7)\geq\frac{1}{2}(2y+z)(2y^7+z^7)\geq y^3z^3(y+z)^2,$$
$$\frac{1}{2}(x+y+z)(x^7+y^7+z^7)\geq\frac{1}{2}(x+2z)(x^7+2z^7)\geq x^3z^3(x+z)^2$$ and by Muirhead 
$$\sum_{cyc}(x-y)^2\left(\frac{1}{2}(x+y+z)(x^7+y^7+z^7)-x^3y^3(x+y)^2\right)\geq$$
$$\geq(x-y)^2\left(\frac{1}{2}(x+y+z)(x^7+y^7+z^7)-x^3y^3(x+y)^2\right)+$$
$$+(x-z)^2\left(\frac{1}{2}(x+y+z)(x^7+y^7+z^7)-x^3z^3(x+z)^2\right)\geq$$
$$=(x-y)^2\left((x+y+z)(x^7+y^7+z^7)-x^3y^3(x+y)^2-x^3z^3(x+z)^2\right)\geq$$
$$\geq(x-y)^2\left((x^3+y^3+z^3)(x^5+y^5+z^5)-x^3y^3(x+y)^2-x^3z^3(x+z)^2\right)=$$
$$=(x-y)^2\left((x^4+y^4+z^4)^2+\sum_{cyc}x^3y^3(x-y)^3-x^3y^3(x+y)^2-x^3z^3(x+z)^2\right)=$$
$$=(x-y)^2((x^4+y^4+z^4)^2+y^3z^3(y-z)^2-4(y^4+z^4)x^4)\geq$$
$$\geq(x-y)^2((x^4+y^4+z^4)^2-4(y^4+z^4)x^4)=(x-y)^2(x^4-y^4-z^4)^2\geq0.$$
Done!
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 :

$$xyz=1 \implies x+y+z\geq 3$$