# For non-negative reals such that $a+b+c\geq x+y+z$, $ab+bc+ca\geq xy+yz+zx$, and $abc\geq xyz$, show $a^k+b^k+c^k\geq x^k+y^k+z^k$ for $0<k<1$

Let $$a$$, $$b$$, $$c$$, $$x$$, $$y$$, $$z$$ be non-negative real numbers such that $$a+b+c \geq x+y+z,$$ $$ab+bc+ca \geq xy+yz+zx,$$ $$abc \geq xyz$$ Show that $$a^k+b^k+c^k \geq x^k+y^k+z^k, \quad 0 < k < 1$$

This is case $$n=3$$ of a problem posted by Ji Chen in the Art of Problem Solving forums, 2008.

I have posted a partial result in an answer below.

I have a proof when $$r = \frac12,$$ for weaker conditons $$a+b+c = x+y+z,$$ $$\min(x, y, z) \leqslant \min(a, b, c),$$ $$\max(a, b, c) \leqslant \max(x, y, z).$$ Indeed, if $$u, v > 0$$ it's easy check $$\sqrt{u} - \sqrt{v} \leqslant \frac{u-v}{2\sqrt{v}}.$$ Assume $$x \geqslant y \geqslant z$$ and $$a \geqslant b \geqslant c$$ then $$x \geqslant a, \; c \geqslant z.$$ Therefore \begin{aligned}\sqrt{x}+\sqrt{y}+\sqrt{z} - \sqrt{a} - \sqrt{b} - \sqrt{c} & \leqslant \frac{x-a}{2\sqrt{a}}+ \frac{y-b}{2\sqrt{b}}+ \frac{z-c}{2\sqrt{c}} \\& \leqslant \frac{x-a}{2\sqrt{b}}+ \frac{y-b}{2\sqrt{b}}+ \frac{z-c}{2\sqrt{c}} \\& =\frac{x+y+z-a-b-c}{2\sqrt{b}}-\frac{(c-z)(\sqrt{b}-\sqrt{c})}{2\sqrt{bc}} \leqslant 0.\end{aligned}

• – lhf
Mar 13, 2020 at 9:28
• I know Newton identity, but how to use, please more detail, thanks
– fzen
Mar 13, 2020 at 9:42
• Why do you say that the proof is simple ? Do you have already one ? Mar 13, 2020 at 12:11
• @fzen I solved your problem. If you want to see my solution show please your attempts. Mar 13, 2020 at 13:31
• @Maximilian Janisch Thank you! Soon it will come. Mar 13, 2020 at 14:39

Let $$a$$, $$b$$ and $$c$$ be different numbers, $$a+b+c=u$$, $$ab+ac+bc=v$$ and $$abc=w$$.

Now, by $$u$$, $$v$$ and $$w$$ we can get any permutations of $$a$$, $$b$$ and $$c$$, but since $$a^k+b^k+c^k$$ is a symmetric expression, we can think that $$a^k+b^k+c^k=f(u,v,w)$$ and we need to prove that

$$f$$ increases as a function of $$u$$, as a function of $$v$$ and as a function of $$w$$.

Id est, it's enough to prove that: $$\frac{\partial f}{\partial u}\geq0,$$ $$\frac{\partial f}{\partial v}\geq0$$ and $$\frac{\partial f}{\partial w}\geq0.$$

Indeed, $$1=\frac{\partial(a+b+c)}{\partial u}=\frac{\partial a}{\partial u}+\frac{\partial b}{\partial u}+\frac{\partial c}{\partial u},$$ $$0=\frac{\partial(ab+ac+bc)}{\partial u}=\frac{\partial a}{\partial u}b+\frac{\partial b}{\partial u}a+\frac{\partial a}{\partial u}c+\frac{\partial c}{\partial u}a+\frac{\partial b}{\partial u}c+\frac{\partial c}{\partial u}b=$$ $$=(b+c)\frac{\partial a}{\partial u}+(a+c)\frac{\partial b}{\partial u}+(a+b)\frac{\partial c}{\partial u}.$$ Also, $$0=\frac{\partial(abc)}{\partial u}=bc\frac{\partial a}{\partial u}+ac\frac{\partial b}{\partial u}+ab\frac{\partial c}{\partial u}.$$ Now, the determinant of this system it's $$\Delta=\left|\left(\begin{array}{cc} 1&1&1\\b+c&a+c&a+b\\bc&ac&ab\end{array}\right)\right|=\sum_{cyc}(ab(a+c)-bc(a+c))=$$ $$=\sum_{cyc}(a^2b-a^2c)=(a-b)(a-c)(b-c);$$ $$\Delta_{\frac{\partial a}{\partial u}}=\left|\left(\begin{array}{cc}1&1&1\\0&a+c&a+b\\0&ac&ab\end{array}\right)\right|=(a+c)ab-ac(a+b)=a^2(b-c),$$ which gives $$\frac{\partial a}{\partial u}=\frac{a^2(b-c)}{(a-b)(a-c)(b-c)}=\frac{a^2}{(a-b)(a-c)}.$$ Similarly, $$\frac{\partial b}{\partial u}=\frac{b^2}{(b-a)(b-c)}$$ and $$\frac{\partial c}{\partial u}=\frac{c^2}{(c-a)(c-b)}.$$ By the same way we can get that $$\left(\frac{\partial a}{\partial v},\frac{\partial b}{\partial v},\frac{\partial c}{\partial v}\right)=\left(-\frac{a}{(a-b)(a-c)},-\frac{b}{(b-a)(b-c)},-\frac{c}{(c-a)(c-b)}\right)$$ and $$\left(\frac{\partial a}{\partial w},\frac{\partial b}{\partial w},\frac{\partial c}{\partial w}\right)=\left(\frac{1}{(a-b)(a-c)},\frac{1}{(b-a)(b-c)},\frac{1}{(c-a)(c-b)}\right).$$ Now, let $$a>b>c$$.

Thus, by the Lagrange's theorem there is $$t\in[b,a],$$ for which $$\frac{a^{k+1}-b^{k+1}}{a-b}=(k+1)t^k\geq(k+1)b^k,$$ which gives $$a^{k+1}\geq b^{k+1}+(k+1)b^k(a-b).$$ Also, there is $$p\in[c,b],$$ for which $$\frac{b^{k+1}-c^{k+1}}{b-c}=(k+1)p^k\leq(k+1)b^k,$$ which says $$c^{k+1}\geq b^{k+1}-(k+1)b^k(b-c).$$ Thus, $$\frac{\partial f}{\partial u}=\sum_{cyc}\frac{\partial f}{\partial a}\frac{\partial a}{\partial u}=\sum_{cyc}\frac{ka^{k-1}\cdot a^2}{(a-b)(a-c)}=k\sum_{cyc}\frac{a^{k+1}}{(a-b)(a-c)}\geq$$ $$\geq k\left(\frac{b^{k+1}+(k+1)b^k(a-b)}{(a-b)(a-c)}+\frac{b^{k+1}}{(b-a)(b-c)}+\frac{b^{k+1}-(k+1)b^k(b-c)}{(c-a)(c-b)}\right)=0.$$ By the same way we can prove that $$\frac{\partial f}{\partial v}\geq0$$ and $$\frac{\partial f}{\partial w}\geq0.$$

Now, since our proof is valid for $$a\rightarrow b^+$$ and for $$b\rightarrow c^+$$ and since $$f$$ is a continues function, we have that $$f$$ increases for any positive $$a$$, $$b$$ and $$c$$, which ends a proof.

• Nice. Could you point out to a reference where it is shown that the function f exists? I know that any symmetric polynomial is a polynomial in the elementary symmetric polynomials but for 0<k<1 the function is not a polynomial. An other questions that I had is if you could explain the meaning of expressions like $\frac{\partial a}{\partial u}$. You are regarding a as a function of what here (a is not a symmetric polynomial) ? Or asked another way, what are you keeping constant when performing the partial derivative ? Thanks and sorry for the questions... Mar 14, 2020 at 12:08
• Thomas, this is a function because for given u, v and w there is an unique set $\{a,b,c\}$ Mar 14, 2020 at 12:28
• Thank you @Michael Rozenberg
– fzen
Mar 14, 2020 at 12:43
• I see. Indeed that is the set of roots of the polynomial $x^3-ux^2+vx-w=0$. Let's suppose that we have $x_1<x_2<x_3$ as roots for a particular set of $u,v,w$. We can have than $a=x_1,b=x_2,c=x_3$ or also $a=x_2,b=x_1,c=x_3$ or every permutation. Are you imposing $a<b<c$ to fix what are called a,b and c individually ? Mar 14, 2020 at 13:29
• Yes this is clear. I was just asking if to define a as a function of u,v, w you need that condition. The condition a>b>c appears only later in the derivation, after you perform the partial derivative, and therefore I wanted to check if I was missing something. I guess more or less I am getting the solution now. thanks! Mar 14, 2020 at 18:19

Thank @Michael Rozenberg, I have a proof when $$k = \frac12,$$ for weaker conditons $$a+b+c = x+y+z,$$ $$\min(x, y, z) \leqslant \min(a, b, c),$$ $$\max(a, b, c) \leqslant \max(x, y, z).$$ Indeed, if $$u, v > 0$$ it's easy check $$\sqrt{u} - \sqrt{v} \leqslant \frac{u-v}{2\sqrt{v}}.$$ Assume $$x \geqslant y \geqslant z$$ and $$a \geqslant b \geqslant c$$ then $$x \geqslant a, \; c \geqslant z.$$ Therefore \begin{aligned}\sqrt{x}+\sqrt{y}+\sqrt{z} - \sqrt{a} - \sqrt{b} - \sqrt{c} & \leqslant \frac{x-a}{2\sqrt{a}}+ \frac{y-b}{2\sqrt{b}}+ \frac{z-c}{2\sqrt{c}} \\& \leqslant \frac{x-a}{2\sqrt{b}}+ \frac{y-b}{2\sqrt{b}}+ \frac{z-c}{2\sqrt{c}} \\& =\frac{x+y+z-a-b-c}{2\sqrt{b}}-\frac{(c-z)(\sqrt{b}-\sqrt{c})}{2\sqrt{bc}} \leqslant 0.\end{aligned}