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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: $xyz\leq 1$.

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    $\begingroup$ Can you tell us why you want to prove these inequalities and how they arose? $\endgroup$
    – David Roberts
    Mar 9, 2021 at 9:57
  • $\begingroup$ Also, maybe wait a little bit between questions. You have asked two in quick succession. $\endgroup$
    – David Roberts
    Mar 9, 2021 at 9:58
  • $\begingroup$ 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 $\endgroup$ Mar 9, 2021 at 11:51
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    $\begingroup$ OK, fair enough. It is good to give the source, if the question is not original to you. :-) $\endgroup$
    – David Roberts
    Mar 9, 2021 at 21:10

6 Answers 6

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

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

notebook 1


Here is a "more human" proof:

Substituting $z=\frac1{xy}$, rewrite your first inequality as \begin{equation} 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 \end{equation} 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):

notebook 2

The latter proof would probably take many thousands pages.

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    $\begingroup$ What does "$y = \infty-$" mean? $\endgroup$
    – LSpice
    Mar 9, 2021 at 15:49
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    $\begingroup$ @LSpice : I have replaced that by "$y>0$ is large enough". $\endgroup$ Mar 9, 2021 at 16:02
  • $\begingroup$ 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? $\endgroup$
    – LSpice
    Mar 9, 2021 at 21:51
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    $\begingroup$ @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. $\endgroup$ Mar 10, 2021 at 1:17
  • $\begingroup$ 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. $\endgroup$ Mar 10, 2021 at 2:30
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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.

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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.

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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 $$(42,0,0)\succ(40,1,1).$$

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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 $0$.

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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 $i+j+k=12$.

mathoverflow 385942

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.

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  • $\begingroup$ The second one is nice. $\endgroup$
    – River Li
    Oct 16 at 1:58

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