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)
On
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$$
On
Let us prove the equivalent inequality $$f(x,y)= \left(x^5y^5+\dfrac1{x^5}+\dfrac1{y^5}\right)^2 -3\left(x^7y^7+\dfrac1{x^7}+\dfrac1{y^7}\right) \ge 0,\quad (x,y)\in\mathbb R_+^2,\tag1$$ the stationary points of which can be defined from the system $f'_x=0,\ f'_y = 0,$ or $$\begin{cases} 10\left(x^4y^5-\dfrac1{x^6}\right) \left(x^5y^5+\dfrac1{x^5}+\dfrac1{y^5}\right) -21\left(x^6y^7-\dfrac1{x^8}\right)=0\\ 10\left(x^5y^4-\dfrac1{y^6}\right) \left(x^5y^5+\dfrac1{x^5}+\dfrac1{y^5}\right) -21\left(x^7y^6-\dfrac1{y^8}\right)=0,\\ \end{cases}\tag2$$ \begin{align} &\left(x^4y^5-\dfrac1{x^6}\right)\left(x^7y^6-\dfrac1{y^8}\right) = \left(x^5y^4-\dfrac1{y^6}\right)\left(x^6y^7-\dfrac1{x^8}\right),\\[4pt] &x^2(x^{10}y^5-1)(x^7y^{14}-1) = y^2(x^5y^{10}-1)(x^{14}y^7-1),\\[4pt] &x^4y^2(1-x^{10}y^5-x^7y^{14}) = y^4x^2(1-x^5y^{10}-x^{14}y^7).\\[4pt] \end{align} Let $$u=x^2y,\quad v=xy^2,$$ then \begin{align} &u^2(1-u^5-v^7) = v^2(1-u^7-v^5),\\ &(u-1)(v-1)(v-u)(u^5v+u^5+u^4v^2+2u^4v+u^4\\ &+u^3v^3+2u^3v^2+2u^3v+u^3+u^2v^4+2u^2v^3+2u^2v^2+2u^2v+u^2\\ &+uv^5+2uv^4+2uv^3+2uv^2+2uv+u+v^5+v^4+v^3+v^2+v) = 0 \end{align} (see also Wolfram Alpha), \begin{align} &(u=1)\vee(v=1)\vee(u=v),\\ &(x^2y=1)\vee(xy^2=1)\vee(x^2y=xy^2),\\ \end{align} $$\left(y=\dfrac1{x^2}\right)\vee\left(x=\dfrac1{y^2}\right)\vee(x=y).\tag3$$
$\mathbf{\text{Case }y=\dfrac1{x^2}}$
Using the second equation of $(2)$ factor $x^8:$ \begin{align} &10(1-x^{15})(x^{15}+2)-21(x^3-x^{24})=0,\tag4\\ &10(u^{10}+u^5-2)-21(u^8-u) = 0,\quad\text{where}\quad u=x^3,\\ &(u-1)P_9(u)=0,\\ &\text{where the polynomial}\\ &P_9(u) = 10u^9+10u^8-11u^7-11u^6-11u^5-u^4-u^3-u^2-u+20\\ \end{align} has the single real root $u\approx -1.57861$ and is positive if $u>0$ (see also Wolfram Alpha).
So the single staionary point in this case is $$\{x,\ y\ ,f(x,y)\}=\{1,\ 1,\ 0\}.\tag5$$
$\mathbf{\text{Case }x=\dfrac1{y^2}}$
This case leads to the solution $(5),$ because the task is symmetric on the variables $x,y.$
$\mathbf{\text{Case }y=x}$
The second equation of $(2)$ factor $x^{11}$ leads to the equation $(4),$ with the same solution $(5).$
$\mathbf{\text{Bounds}}$
Note that $$\lim\limits_{x\to+0}f(x,y) = \lim\limits_{x\to+infty}f(x,y) = \lim\limits_{y\to+0}f(x,y) = \lim\limits_{y\to+\infty}f(x,y) = +\infty\ge0.$$
Therefore, $$\boxed{\mathbf{(a^5+b^5+c^5)^2\ge3(a^7+b^7+c^7).}}$$
We can use the Vasc's EV-Method, Corollary 1.8 (b):
https://www.emis.de/journals/JIPAM/images/059_06_JIPAM/059_06.pdf
Indeed, we need to prove that: $$(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!