prove this inequality by $abc=1$

Solution 1:

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!