Olympiad Inequality $\sum\limits_{cyc} \frac{x^4}{8x^3+5y^3} \geqslant \frac{x+y+z}{13}$

Solution 1:

A big problem we get around $(x,y,z)=(0.822,1.265,1.855)$.

The Buffalo Way helps:

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

Hence, $\frac{13}{5}\prod\limits_{cyc}(8x^3+5y^3)\left(\sum\limits_{cyc}\frac{x^4}{8x^3+5y^3}-\frac{x+y+z}{13}\right)=$

$$=156(u^2-uv+v^2)x^8+6(65u^3+189u^2v-176uv^2+65v^3)x^7+$$ $$+2(377u^4+1206u^3v+585u^2v^2-1349uv^3+377v^4)x^6+$$ $$+3(247u^5+999u^4v+1168u^3v^2-472u^2v^3-726uv^4+247)x^5+$$ $$+3(117u^6+696u^5v+1479u^4v^2+182u^3v^3-686u^2v^4-163uv^5+117v^6)x^4+$$ $$+(65u^7+768u^6v+2808u^5v^2+2079u^4v^3-1286u^3v^4-585u^2v^5+181uv^6+65v^7)x^3+$$$$+3uv(40u^6+296u^5v+472u^4v^2-225u^2v^4+55uv^5+25v^6)x^2+ $$ $$+u^2v^2(120u^5+376u^4v+240u^3v^2-240u^2v^3-25uv^4+75v^5)x+$$ $$+5u^3v^3(8u^4+8u^3v-8uv^3+5v^4)\geq$$ $$\geq u^5v^5(156t^8+531t^7+2t^6-632t^5-152t^4+867t^3+834t^2+299t+40)\geq0$$

Done!

For example, we'll prove that $$6(65u^3+189u^2v-176uv^2+65v^3)\geq531\sqrt{u^3v^3},$$ which gives a coefficient $531$ before $t^7$ in the polynomial $156t^8+531t^7+2t^6-632t^5-152t^4+867t^3+834t^2+299t+40.$

Indeed, let $u=k^2v$, where $k>0$.

Thus, we need to prove that: $$130k^6+378k^4-177k^3-352k^2+130\geq0$$ and by AM-GM we obtain: $$130k^6+378k^4-177k^3-352k^2+130=$$ $$=130\left(k^3+\frac{10}{13}k-1\right)^2+\frac{k}{13}(2314k^3+1079k^2-5576k+2600)\geq$$ $$\geq\frac{k}{13}\left(8\cdot\frac{1157}{4}k^3+5\cdot\frac{1079}{5}k^2+21\cdot\frac{2600}{21}-5576k\right)\geq$$ $$\geq\frac{k^2}{13}\left(34\sqrt[34]{\left(\frac{1157}{4}\right)^8\left(\frac{1079}{5}\right)^5\left(\frac{2600}{21}\right)^{21}}-5576\right)>0.$$ We'll prove that $$ 2(377u^4+1206u^3v+585u^2v^2-1349uv^3+377v^4)\geq2u^2v^2,$$ for which it's enough to prove that: $$377t^4+1206t^3+584t^2-1349t+377\geq0$$ or $$t^4+\frac{1206}{377}t^3+\frac{584}{377}t^2-\frac{1349}{377}t+1\geq0$$ or $$\left(t^2+\frac{603}{377}t-\frac{28}{29}\right)^2+\frac{131015t^2-69589t+9633}{142129}\geq0,$$ which is true because $$69589^2-4\cdot131015\cdot9633<0.$$

Solution 2:

This is a question of the symmetric type, such as listed in:

• Why does Group Theory not come in here?

With a constraint $\;x+y+z=1\;$ and $\;x,y,z > 0$ . Sort of a general method to transform such a constraint into the inside of a triangle in 2-D has been explained at length in:

• How prove this inequality $(a^2+bc^4)(b^2+ca^4)(c^2+ab^4) \leq 64$

Our function $f$ in this case is: $$ f(x,y,z) = \frac{x^4}{8x^3+5y^3}+\frac{y^4}{8y^3+5z^3}+\frac{z^4}{8z^3+5x^3} - \frac{1}{13} $$ And the minimum of that function inside the abovementioned triangle must shown to be greater or equal to zero. Due to symmetry - why oh why can it not be proved with Group Theory - an absolute minimum of the function is expected at $(x,y,z) = (1/3,1/3,1/3)$. Another proof without words is attempted by plotting a contour map of the function, as depicted. Levels (nivo) of these isolines are defined (in Delphi Pascal) as:

nivo := min + sqr(g/grens)*(max-min); { sqr = square ; grens = 20 ; g = 0..grens }

The whiteness of the isolines is proportional to the (positive) function values; they are almost black near the minimum and almost white near the maximum values. Maximum and minimum values of the function are observed to be:

 0.00000000000000E+0000 < f < 4.80709198767699E-0002

The little $\color{blue}{\mbox{blue}}$ spot in the middle is where $\,0 \le f(x,y,z) < 0.00002$ .