# how to prove this high degree inequality

Let $$x$$,$$y$$,$$z$$ be positive real numbers which satisfy $$xyz=1$$. Prove that: $$(x^{10}+y^{10}+z^{10})^2 \geq 3(x^{13}+y^{13}+z^{13})$$.

And there is a similar question: Let $$x$$,$$y$$,$$z$$ be positive real numbers which satisfy the inequality $$(2x^4+3y^4)(2y^4+3z^4)(2z^4+3x^4) \leq(3x+2y)(3y+2z)(3z+2x)$$. Prove this inequality: $$xyz\leq 1$$.

• Can you tell us why you want to prove these inequalities and how they arose? Mar 9, 2021 at 9:57
• Also, maybe wait a little bit between questions. You have asked two in quick succession. Mar 9, 2021 at 9:58
• thank you for your question,I asked the two question just out of curosity, neither for examination nor profit propose.I met this two questions online,here is the link, and I translate it to English. mp.weixin.qq.com/s/w9_isyUk5ie0Oo69c7g9lg Mar 9, 2021 at 11:51
• OK, fair enough. It is good to give the source, if the question is not original to you. :-) Mar 9, 2021 at 21:10

These inequalities are algebraic and thus can be proved purely algorithmically.

Mathematica takes a minute or two for this proof of your first inequality:

Here is a "more human" proof:

Substituting $$z=\frac1{xy}$$, rewrite your first inequality as $$$$f(x,y)\mathrel{:=}\left(\frac{1}{x^{10} y^{10}}+x^{10}+y^{10}\right)^2-3 \left(\frac{1}{x^{13} y^{13}}+x^{13}+y^{13}\right)\ge0$$$$ and then as \begin{align} g(x,y)&\mathrel{:=}f(x,y)x^{20} y^{20} \\ &=x^{40} y^{20}-3 x^{33} y^{20}+2 x^{30} y^{30}+x^{20} y^{40}-3 x^{20} y^{33} \\ &+2 x^{20} y^{10}+2 x^{10} y^{20}-3 x^7 y^7+1\ge0, \end{align} for $$x,y>0$$.

Further, \begin{align} g_1(x,y)\mathrel{:=}{}\frac{g'_x(x,y)}{x^6 y^7}&=40 x^{33} y^{13}-99 x^{26} y^{13}+60 x^{23} y^{23}+20 x^{13} y^{33} \\ &-60 x^{13} y^{26}+40 x^{13} y^3+20 x^3 y^{13}-21, \\ g_2(x,y)\mathrel{:=}\frac{g'_y(x,y)}{x^7 y^6}&=20 x^{33} y^{13}-60 x^{26} y^{13}+60 x^{23} y^{23}+40 x^{13} y^{33} \\ &-99 x^{13} y^{26}+20 x^{13} y^3+40 x^3 y^{13}-21, \end{align} and the only positive roots of the resultants of $$g_1(x,y)$$ and $$g_2(x,y)$$ with respect to $$x$$ and $$y$$ are $$y=1$$ and $$x=1$$, respectively. So, $$(1,1)$$ is the only critical point of $$g$$.

Next, all the coefficients of the polynomial $$g(1+u,1+v)$$ in $$u,v$$ are nonnegative. Therefore and because of the symmetry $$x\leftrightarrow y$$, it remains to consider the cases (i) $$0\le x\le1$$ and $$y>0$$ is large enough and (ii) $$x=0$$.

For case (i), we have $$g(x,y)\ge1 - 3 x^7 y^7 + 2 x^{10} y^{20}>0$$. For case (ii), we have $$g(0,y)=1>0$$.

So, your first inequality is proved, again.

It took Mathematica about 1.8 hours to prove your second inequality (click on the image to enlarge it):

The latter proof would probably take many thousands pages.

• What does "$y = \infty-$" mean? Mar 9, 2021 at 15:49
• @LSpice : I have replaced that by "$y>0$ is large enough". Mar 9, 2021 at 16:02
• Thanks! Just so that it's here in fulltext: the first Mathematica command is Reduce[(x^(10)+y^(10)+z^(10))^2<3(x^(13)+y^(13)+z^(13))&&x>0&&y>0&&z>0&&x y z>1,{x,y,z},Reals], and the second is Reduce[(2x^4+3y^4)(3x^4+2z^4)(2y^4+3z^4)<=(3x+2y)(3y+2z)(2z+3z)&&x>0&&y>0&&z>0&&x y z>1,{x,y,z},Reals]//AbsoluteTiming. Since the inequalities seem to be true (at least, you prove the first one), what does it mean that Mathematica returns False? Mar 9, 2021 at 21:51
• @LSpice : Both "inequalities" are actually implications of the form $C:=(A\implies B)$. In both cases, I asked Mathematica to Reduce[] the negation $\neg C=(A\ \&\ \neg B)$ of the implication $C$ (where $\neg$ is the negation symbol, $A$ is an equality or an inequality, and $B$ is an inequality). In both cases, Mathematica Reduce[]'d the negation $\neg C$ of $C$ to False. So, $C$ is true. Mar 10, 2021 at 1:17
• thank you for your detailed solution and explaination. the application of calculus is awesome.it's so amzing to use Mathematica to prove or disprove inequality.Never have I realised that this software can be so powerful. Mar 10, 2021 at 2:30

Here is a little less computer-assisted approach to both inequalities than the one suggested by Iosif Pinelis. Namely, I use the properties of rational one-variable functions which are seen from their graphs without thinking about how to prove them rigorously.

1. Fix $$xyz=1$$ and $$x^{10}+y^{10}+z^{10}:=S$$. Look for a maximum of $$x^{13}+y^{13}+z^{13}$$. It is achieved (the set of admissible triples is compact), and at the maximum points the gradients of $$xyz,x^{10}+y^{10}+z^{10},x^{13}+y^{13}+z^{13}$$ are linearly dependent, that is, we should have $$\alpha\cdot(yz,xz,xy)+\beta\cdot 10(x^9,y^9,z^9)+\gamma\cdot 13(x^{12},y^{12},z^{12})=0$$ where the real coefficients $$(\alpha,\beta,\gamma)$$ are not simultaneously zero. This means that all numbers $$x,y,z$$ solve the same equation $$f(t):=a+bt^{10}+ct^{13}=0$$, where $$a=\alpha xyz$$, $$b=10\beta$$, $$c=13\gamma$$. Such an equation may have at most two different positive solutions by Descartes' rule of signs. That is, two of $$x,y,z$$ must be equal. Without loss of generality we may assume that $$y=x$$, then $$z=1/x^2$$ and we should prove a 1-variable inequality $$(2x^{10}+x^{-20})^2\geqslant 3(2x^{13}+x^{-26}).$$ I do not see any nice explanation why this is true, but looking at the graphs we see that the ratio of exponents 13:10 may be increased to approximately 1.4047 (you may see that this graph is above the x-axis, but for the value of parameter 1.4048 it crosses it already.

2. Denote $$a=y/x,b=z/y,c=x/z$$. Then $$abc=1$$ and we are given $$(xyz)^4(2+3a^4)(2+3b^4)(2+3c^4)\leqslant xyz(3+2a)(3+2b)(3+2c)$$. Thus for establishing $$xyz\leqslant 1$$ it suffices (and is actually necessary) to check that $$(3+2a)(3+2b)(3+2c)\leqslant (2+3a^4)(2+3b^4)(2+3c^4) \quad (1)$$ whenever $$a,b,c$$ are positive numbers with $$abc=1$$. (1) is equivalent to $$F(a,b,c):=h(a)+h(b)+h(c)\geqslant 0, \,\, \text{where}\,\, h(x):=\log(2+3x^4)-\log(3+2x).$$ First of all, I claim that $$F$$ attains its minimal value on the set $$\Omega:=\{(a,b,c):abc=1, a,b,c>0\}$$. Indeed, consider a sequence $$(a_n,b_n,c_n)$$ for which $$F(a_n,b_n,c_n)$$ approaches the infimum of $$F$$ on $$\Omega$$. Since $$h(x)$$ is bounded from below on $$(0,\infty)$$ and tends to $$+\infty$$ for large $$x$$, we conclude that $$a_n,b_n,c_n$$ must be bounded, then we may choose a convergent subsequence and get a minimizer. So, denote the minimizer by $$(a_0,b_0,c_0)$$. the gradients of $$F(a,b,c)$$ and $$abc$$ at the point $$(a_0,b_0,c_0)$$ must be linearly dependent, thus we may write $$h'(a_0)=\lambda b_0c_0$$, $$h'(b_0)=\lambda a_0c_0$$, $$h'(c_0)=\lambda b_0a_0$$ for certain real $$\lambda$$. In other words, the function $$g(x):=xh'(x)$$ takes the same value at points $$a_0,b_0,c_0$$. Looking at the plot of $$g(x)$$ for positive $$x$$ we see that it takes each value at most twice. Thus two of three variables $$a_0,b_0,c_0$$ must be equal. Without loss of generality $$a_0=b_0=:x$$, $$c_0=1/x^2$$, and we should prove a 1-variable polynomial inequality $$(2+3x^4)^2(2+3/x^8)\geqslant (3+2x)^2(3+2/x^2)\quad \text{for}\quad x>0.$$ Well, it follows from factorization.

Another way. By my previous post it's enough to prove that: $$(x^5+y^5+z^5)^2\geq3xyz(x^7+y^7+z^7)$$ for positives $$x$$, $$y$$ and $$z$$.

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

Thus, by the Vasc's EV Method (see here: https://www.emis.de/journals/JIPAM/images/059_06_JIPAM/059_06.pdf Corollary 1.8(b)) it's enough to prove the last inequality (because it's homogeneous) for $$y=z=1$$, which gives $$(x^5+2)^2\geq3x(x^7+2)$$ or $$(x-1)^2(x^8+2x^7-2x^5-4x^4-2x^3+2x+4)\geq0$$ and the rest is smooth.

We'll prove a stronger inequality:

$$\left(a^{10}+b^{10}+c^{10}\right)^2\geq3(a^{14}+b^{14}+c^{14}),$$ where $$abc=1,$$

for which it's enough to prove that $$(x^5+y^5+z^5)^2\geq3xyz(x^7+y^7+z^7)$$ for positives $$x$$, $$y$$ and $$z$$.

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

Hence, $$\left(x^5 + y^5 + z^5\right)^2-3xyz\left(x^7 +y^7 + z^7\right)=$$ $$=(u^2-uv+v^2)x^8+8(-u^3+2u^2v+2uv^2-v^3)x^7+$$ $$+4(5u^4-17u^3v+50u^2v^2-17uv^3+5v^4)x^6+$$ $$+8(11u^5-20u^4v+25u^2v^2+25u^2v^3-20uv^4+11v^5)x^5+$$ $$+2(63u^6-79u^5v+50u^4v^2+100u^3v^3+50u^2v^4-79uv^5+63u^6)x^4+$$ $$+4(24u^7-21u^6v+5u^5v^2+25u^4v^3+25u^3v^4+5u^2v^5-21uv^6+24v^7)x^3+$$ $$+2(21u^8-12u^7v+10u^5v^3+25u^4v^4+10u^3v^5-12uv^7+21v^8)x^2+$$ $$+(10u^9-3u^8v+10u^5v^4+10u^4v^5-3uv^8+10v^9)x+(u^5+v^5)^2\geq$$ $$\geq\left((u^2-uv+v^2)x^2-8(u+v)(u^2-3uv+v^2)x+4(5u^4-17u^3v+50u^2v^2-17uv^3+5v^4)\right)x^6.$$

Let $$u^2+v^2=tuv$$.

Hence, $$t\geq2$$ and it remains to prove that $$(t-1)(5t^2-17t+40)-4(t+2)(t-3)^2\geq0,$$ which is true because $$(t-1)(5t^2-17t+40)-4(t+2)(t-3)^2=$$ $$=t^3-6t^2+69t-112=t(t-3)^2+60t-112\geq0.$$ Id est, it's enough to prove that $$x^{14}+y^{14}+z^{14}\geq x^{13}+y^{13}+z^{13},$$ where $$x$$, $$y$$ and $$z$$ are positives such that $$xyz=1$$, or $$x^{42}+y^{42}+z^{42}\geq (x^{39}+y^{39}+z^{39})xyz,$$ which is true by Muirhed because $$(42,0,0)\succ(40,1,1).$$

This is to complement the nice answer by Fedor Petrov by a calculus proof of
the inequality $$(2 x^{10} + x^{-20})^2\ge3 (2 x^{13} + x^{-26}),$$ for real $$x>0$$.

Rewrite this inequality as $$f(x):=4 x^{60}-6 x^{53}+4 x^{30}-3 x^{14}+1\ge0.$$ Let $$f_1(x):=\frac{f'(x)}{6x^{13}}=40 x^{46}-53 x^{39}+20 x^{16}-7,\\f_2(x):=\frac{f_1'(x)}{x^{15}}=1840 x^{30}-2067 x^{23}+320,\\ f_3(x):=\frac{f_2'(x)}{69x^{22}}=800 x^7-689.$$ Then, clearly, $$f_2$$ attains its minimum (on $$[0,\infty)$$) at $$(689/800)^{1/7}$$, and this minimum is $$24.7\ldots>0$$. So, $$f_2>0$$ and hence $$f_1$$ is increasing, from $$f_1(0)=-7<0$$ to $$f_1(\infty-)=\infty>0$$. Also, $$f_1(1)=0$$. So, $$f_1\le0$$ on $$[0,1]$$ and $$f_1\ge0$$ on $$[1,\infty)$$. So, $$f$$ attains its minimum (on $$[0,\infty)$$) at $$1$$, and this minimum is $$0$$.

For the first inequality, let us write $$x=X^3$$, $$y=Y^3$$, $$z=Z^3$$, and pass to a homogenized version of it, obtained by multiplying the R.H.S. with $$(xyz)^{7/3}=(XYZ)^7$$. So we have to show: $$(X^{30}+Y^{30}+Z^{30})^2 \ge 3 (X^{39}+Y^{39}+Z^{39})X^7Y^7Z^7\ .$$ Now divide by $$Z^{60}$$ to dehomogenize, so it is enough to show the above for $$Z=1$$ and arbitrary $$X,Y\ge 0$$. Consider the difference function $$f(X,Y)= (X^{30}+Y^{30}+1)^2 - 3 (X^{39}+Y^{39}+1)X^7Y^7\ .$$ Since $$f(1,1)=0$$ and on the boundary $$XY=0$$ we have $$f(X,Y)\ge 1$$, a global minimal value is a local minimum in the domain $$Y,Y>0$$. We search such local minima, they are among the solutions of $$f'(X,Y)=0$$, which leads to the algebraic system of equations \left\{ \begin{aligned} 2(X^{30}+Y^{30}+1)\cdot 30X^{29} &= 3 (46X^{45}Y^7 + 7X^6Y^{46}+7X^6Y^7)\ ,\\ 2(X^{30}+Y^{30}+1)\cdot 30Y^{29} &= 3 (46Y^{45}X^7 + 7Y^6X^{46}+7Y^6X^7)\ . \end{aligned} \right. Now multiply the first equation with $$X$$, the second with $$Y$$. This leads to $$2(X^{30}+Y^{30}+1)\cdot 30X^{30} = 3 (46X^{46}Y^7 + 7X^7Y^{46}+7X^7Y^7) = 2(X^{30}+Y^{30}+1)\cdot 30Y^{30} \ .$$ So a solution must satisfy $$X=Y$$. We plug in this into the above, get $$2(2X^{30}+1)\cdot 30X^{30} = 3 (53X^{53}+7X^{14}) \ .$$ It turns out that $$X=1$$ is the only real positive root. So $$(X,Y)=(1,1)$$ is the only critical point. So it is the point where $$f$$ is globaly minimal.

The second point follows from the inequality $$(2x^4+3y^4)(2y^4+3z^4)(2z^4+3x^4) \ge (3x+2y)(3y+2z)(3z+2x)x^3y^3z^3\ .$$ Let us show the above. After we expand, it is enough to show the domination $$12 \, \sum x^{8} y^{4} + 18 \sum \, y^{8} x^{4} \ge 18 \, \sum x^{5} y^{4} z^{3} + 12 \sum\, x^{5} y^{3} z^{4} \ .$$ To have a better view, let us place the points $$(i,j,k)$$ corresponding to the monomials $$x^iy^jz^k$$ that occur in the plane $$i+j+k=12$$.

Now, a simple human scheme of domination can be found, for instance exploiting: \begin{aligned} 12(5,3,4) &= 5(8,4,0)+ 2(0,8,4) + 5(4,0,8)\ ,\\ 12(5,4,3) &= 5(4,8,0)+ 2(0,4,8) + 5(8,0,4)\ . \end{aligned} Now using the Jensen-convexity of the logarithm in the form $$au+bv+cw\ge u^av^bw^c$$ for $$u,v,w>0$$ and weights $$a,b,c>0$$ with total sum $$a+b+c=1$$, for the above suggested weights $$a=c=5/12$$, $$b=2/12$$, and the monomials $$u=x^8y^4$$, $$v=y^8z^4$$, $$w=z^8x^4$$ for the first line, then the reversed version for the second line, gives \begin{aligned} 12\,x^5y^3z^4 &\le 5\,x^8y^4 + 2\,y^8z^4 + 5\,z^8x^4\ ,\\ 12\,x^5y^4z^3 &\le 5\,x^4y^8 + 2\,y^4z^8 + 5\,z^4x^8\ . \end{aligned}

This concludes the needed domination.