How to prove this high-degree inequality

Question

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: $xyz\leq 1$.

Answer 1

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

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

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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.

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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.

Answer 2

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.

Answer 3

Here is a simple proof:

By Chebyshev sum inequality and AM-GM, we have \begin{align*} x^{13}x^{1/3} + y^{13}y^{1/3} + z^{13}z^{1/3} &\ge \frac13(x^{13} + y^{13} + z^{13})(x^{1/3} + y^{1/3} + z^{1/3})\\ &\ge \frac13(x^{13} + y^{13} + z^{13})\cdot 3\sqrt[3]{x^{1/3}y^{1/3}z^{1/3}}\\ &= x^{13} + y^{13} + z^{13}. \tag{1} \end{align*}

Then, it suffices to prove that $$(x^{10}+y^{10}+z^{10})^2 \geq 3(x^{13+\frac13}+y^{13 + \frac13}+z^{13+ \frac13}). \tag{2}$$

Letting $x = a^{3/10}, y = b^{3/10}, z = c^{3/10}$, it suffices to prove that, for all $a, b, c > 0$ with $abc = 1$, $$(a^3 + b^3 + c^3)^2 \ge 3(a^4 + b^4 + c^4). \tag{3}$$

We have \begin{align*} &(a^3 + b^3 + c^3)^2 - 3(a^4 + b^4 + c^4)\\ ={}& a^6 + b^6 + c^6 + 2a^3b^3 + 2b^3c^3 + 2c^3a^3 - 3(a^4 + b^4 + c^4)\\ \ge{}& a^6 + b^6 + c^6 + (3a^2b^2 - 1) + (3b^2c^2 - 1) + (3c^2a^2 - 1) - 3(a^4 + b^4 + c^4) \tag{4}\\ ={}& a^6 + b^6 + c^6 + 3a^2b^2 + 3b^2c^2 + 3c^2a^2 - 3a^4 - 3b^4 - 3c^4 - 3\\ ={}& a^6 + b^6 + c^6 + 3a^2b^2 + 3b^2c^2 + 3c^2a^2 - 3a^4 - 3b^4 - 3c^4 - 3a^2b^2c^2\\ ={}& (a^2 + b^2 + c^2 - 3)(a^4 + b^4 + c^4 -a^2b^2 - b^2c^2 - c^2a^2)\\ \ge{}& 0 \end{align*} where we use $2a^3b^3 - (3a^2b^2 - 1) = (2ab + 1)(ab - 1)^2 \ge 0$ in (4).

We are done.

Answer 4

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.

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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.

Answer 5

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 positive $x$, $y$ and $z$.

Indeed, let $x^5+y^5+z^5=\text{constant}$ and $x^7+y^7+z^7=\text{constant}$.

Thus, by the Vasc's EV Method (see here: Cîrtoaje - The equal variable method 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.

Answer 6

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 $0$.

Answer 7

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

Answer 8

This is to give an alternative proof of the inequality in Fedor Petrov's nice answer $$(2 x^{10} + x^{-20})^2\ge3 (2 x^{13} + x^{-26})$$ for all $x > 0$.

We have \begin{align*} &2x^{13+1/3} + x^{-26-2/3} - (2x^{13} + x^{-26})\\ ={}& 2(x^{13+1/3} - x^{13}) + (x^{-13-1/3} - x^{-13}) (x^{-13-1/3} + x^{-13})\\ ={}& x^{13}(x^{1/3}-1)(2 - x^{-119/3} - x^{-118/3})\\ \ge{}& 0. \end{align*}

Thus, it suffices to prove that $$(2 x^{10} + x^{-20})^2\ge 3(2x^{13+1/3} + x^{-26-2/3}).$$

Letting $x = a^{3/10}$, it suffices to prove that, for all $a > 0$, $$(2a^3 + a^{-6})^2 \ge 3(2a^4 + a^{-8}).$$

We have \begin{align*} &(2a^3 + a^{-6})^2 - 3(2a^4 + a^{-8})\\ ={}& 4a^6 + 4a^{-3}+ a^{-12} - 6a^4 - 3a^{-8}\\ \ge\,& 4a^6 + 2(3a^{-2} - 1) + a^{-12} - 6a^4 - 3a^{-8}\\ ={}& \frac{(2a^2+1)(2a^{12} - 2a^6 + 1)(a^2-1)^2}{a^{12}}\\ \ge{}& 0 \end{align*} where we have used $2a^{-3} - (3a^{-2} - 1) = \frac{(a+2)(a-1)^2}{a^3} \ge 0$.

We are done.