AoPS - Problem

mihaig

06.04.2019 19:21

Solving (partially) this outstanding problems,I created two problems. I hope the readers we'll like them and I also hope that they shall help the reader to finish the main task.

Attachments:

Pathological

07.04.2019 00:32

We claim that the answer is all $n \ge 0$. For $n < 0$, $$(a, b, c) = (2, 3, 4), (x, y, z) = (3.5, \frac{5.5 + \sqrt{5.5^2 - 4*48/7}}{2}, \frac{5.5 - \sqrt{5.5^2 - 4*48/7}}{2})$$show that $a^n + b^n + c^n \ge x^n + y^n + z^n$ is not necessarily true. Now, we'll show that all $n \ge 0$ work. Let $p = abc, s = a+b+c$. To prove this, we'll first show the following lemma: $\mathbf{Lemma}$ $\mathbf{1.}$ The set of positive reals $r$ for which there exist $a, b, c \in \mathbb{R}^{+}$ so that $abc =p$, $a + b + c = s$, and $r = \max(a, b, c)$ is a closed interval. ProofLet this set of positive reals $r$ be $S$. Notice that $a$ uniquely determines $b, c$ as: $$b , c= \frac{(s-a) \pm \sqrt{(s-a)^2 - \frac{4p}{a}}}{2}.$$Hence, $r \in S$ is equivalent to $(s-r)^2 \ge \frac{4p}{r}$ and $(s-r) + \sqrt{(s-r)^2 - \frac{4p}{r}} \le 2r.$ We clearly have that $S \subseteq [\frac{s}{3}, s).$ Let $T$ be the set of positive reals $r$ in $[\frac{s}{3}, s)$ for which $(s-r)^2 \ge \frac{4p}{r} \Leftrightarrow r(s-r)^2 \ge 4p$. Since $r(s-r)^2$ is decreasing on $[\frac{s}{3}, s),$ it's clear that $T$ is an interval of the form $[\frac{s}{3}, t]$ for some $t \in [\frac{s}{3}, s)$. Now, the second condition is equivalent to: $$\sqrt{(s-r)^2 - \frac{4p}{r}} \le 3r-s \Leftrightarrow (s-r)^2 - \frac{4p}{r} \le (s-3r)^2,$$ where we used that $r \in [\frac{s}{3}, s)$. Upon expanding, the above is equivalent to $p \ge r^2(s-2r)$, so the subset of $T$ satisfying this is clearly an interval since $r^2(s-2r)$ is decreasing on $[\frac{s}{3}, s).$ Hence, the claim is proven. $\blacksquare$ Let $S$ be the set of possible values attained by $\max(a, b, c)$ as $a, b, c \in \mathbb{R}^{+}$ vary while satisfying $abc =p, a+b+c = s.$ $\mathbf{Lemma}$ $\mathbf{2.}$ As $r$ varies in $S$, define $f(r)$ as the value of $ab + bc + ca$ when $a = r$, and $a+b+c = s, abc = p$. Then $r$ is nonincreasing on $S$. ProofNote that $ab+bc+ca = r(b+c) + bc = r(s-r) + \frac{p}{r}.$ Hence, since $S$ is an interval, it only suffices to show that $s - 2r - \frac{p}{r^2} = f'(r) \le 0.$ To show this, we require that $r + r + \frac{p}{r^2} \ge s.$ However, since $b, c \in [\frac{p}{r^2}, r]$ and $bc = r \cdot \frac{p}{r^2},$ it's clear that $b + c \le r + \frac{p}{r^2}$ (say, by factoring the above), and so hence $r + r + \frac{p}{r^2} \ge r + b + c = s,$ as desired. $\blacksquare$ Let's write $d_i = a^i + b^i + c^i$ and $w_i = x^i + y^i + z^i$ for all $ i\in \mathbb{Z}.$ Set $t_i = d_i - w_i$. Observe that: $$d_n = (a+b+c)d_{n-1} - (ab+bc+ca)d_{n-2} + abcd_{n-3} \ge (x+y+z)d_{n-1} - (xy+yz+zx)d_{n-2} + xyzd_{n-3},$$where we used Lemma $2$. From this and a similar expression for $w_n$, we obtain that $t_n \ge (x+y+z) t_{n-1} - (xy+yz+zx)t_{n-2} + xyz t_{n-3}.$ $(1)$ Notice that $t_0 = t_1 = 0$ and $t_2 = 2(xy+yz+zx-ab-bc-ca) \ge 0$ by Lemma $2$. We claim that these values and the inequality above are enough to imply that $t_n \ge 0, \forall n\in \mathbb{N}.$ Define the sequence $\{h_n\}$ by $h_0 = h_1 = 0$, $h_2 = 1$, and $h_{n+3} = (x+y+z)h_{n+2} - (xy+yz+zx)h_{n+1} + xyzh_n$ for $n \ge 0.$ $\mathbf{Claim.}$ $h_i \ge 0, \forall i \in \mathbb{N}.$ ProofSuppose that $x \ge y \ge z$. Let $P_i(x, y, z)$ denote the polynomial $\frac{x^i(y-z) + y^i(z-x) + z^i(x-y)}{(x-y)(y-z)(x-z)}$ (note that the denominator divides the numerator since $x = y, y=z, z=x$ would make the numerator equal to zero). Then, it's easy to see that $h_i = P_i(x, y, z).$ It therefore suffices to prove that $P_i(x, y, z) \ge 0$ for $i \in \mathbb{N}$. Since $P_i$ is clearly continuous in $x, y, z$, we only need to show that $P_i(x, y, z) \ge 0$ when $x>y>z$, since if two of them were equal we can just take the limit as one approaches the other. In that case, it suffices to prove that $x^i(y-z) + y^i(z-x) + z^i(x-y) \ge 0$ when $x > y > z$. In fact, we will show that $x^i(y-z) + y^i(z-x) + z^i(x-y) \ge 0$ as long as $x \ge y \ge z$, which clearly implies that it's true for $x>y>z.$ Let's fix $y, z$. Now, notice that the derivative of the expression with respect to $x$ is $ix^{i-1}(y-z) -y^i + z^i = (y-z) (ix^{i-1} - y^{i-1} - y^{i-2}z - \cdots - yz^{i-2} - z^{i-1}) \ge 0$, where we used the fact that both multiplicands are $\ge 0$. Since we've shown that it's nondecreasing in $x$, and that $x \ge y$, it suffices to verify it when $x = y$. However, it's clear that equality holds in the $x = y$ case, and so the claim is proven. $\blacksquare$ With the claim, we are now ready to show that $t_n \ge 0$ for all $n \in \mathbb{N}.$ We will show by induction on $n+2c$ that: $$t_{n+c} \ge h_{c+2} t_n + (h_{c+3} - (x+y+z)h_{c+2})t_{n-1} + xyzh_{c+1}t_{n-2}$$ for all $n \ge 2, c \ge 1.$ Call the above equation $(2)$. When $c = 1$, it is simply given in $(1)$. Assume it holds for $n+2c \le k$. When $n+2c = k+1,$ we wish to show that: $$t_{n+c} \ge h_{c+2} t_n + (h_{c+3} - (x+y+z)h_{c+2})t_{n-1} + xyzh_{c+1}t_{n-2}.$$Assume that $c > 1$, since otherwise we're done. By the inductive hypothesis, since $(n+1) + 2(c-1) \le k$, we know that: $$t_{n+c} \ge h_{c+1}t_{n+1} + (h_{c+2} - (x+y+z)h_{c+1})t_n + xyzh_ct_{n-1}.$$ However, since $h_{c+1} \ge 0$ by the claim, we have that: $$RHS \ge^{(1)} h_{c+1} ((x+y+z)t_n - (xy+yz+zx)t_{n-1} + xyzt_{n-2}) + (h_{c+2} - (x+y+z)h_{c+1})t_n + xyzh_ct_{n-1} = $$$$h_{c+2}t_n + (xyzh_c - (xy+yz+zx)h_{c+1})t_{n-1} + xyzh_{c+1} t_{n-2}=$$$$h_{c+2}t_n + (h_{c+3} - (x+y+z)h_{c+2})t_{n-1} + xyzh_{c+1}t_{n-2},$$ which is exactly what we wanted. Therefore, the induction is complete, and $(2)$ is proven. By letting $n = 2$, we obtain that for all $c \ge 1$ $t_{c+2} \ge h_{c+2}t_2 \ge 0$, where we used that $h_{c+2}, t_2 \ge 0.$ Therefore, we've shown that $t_{n} \ge 0$ for all $n \ge 3$, and as $t_0, t_1, t_2 \ge 0$ are known, we've shown that $t_n \ge 0$ for all $n \in \mathbb{N}$, which clearly solves the problem. In conclusion, the answer is all $n \ge 0$. $\square$

hellomath010118

17.04.2019 13:52

Pathological wrote:

We claim that the answer is all $n \ge 0$. For $n < 0$, $$(a, b, c) = (2, 3, 4), (x, y, z) = (3.5, \frac{5.5 + \sqrt{5.5^2 - 4*48/7}}{2}, \frac{5.5 - \sqrt{5.5^2 - 4*48/7}}{2})$$show that $a^n + b^n + c^n \ge x^n + y^n + z^n$ is not necessarily true. Now, we'll show that all $n \ge 0$ work. Let $p = abc, s = a+b+c$. To prove this, we'll first show the following lemma: $\mathbf{Lemma}$ $\mathbf{1.}$ The set of positive reals $r$ for which there exist $a, b, c \in \mathbb{R}^{+}$ so that $abc =p$, $a + b + c = s$, and $r = \max(a, b, c)$ is a closed interval. ProofLet this set of positive reals $r$ be $S$. Notice that $a$ uniquely determines $b, c$ as: $$b , c= \frac{(s-a) \pm \sqrt{(s-a)^2 - \frac{4p}{a}}}{2}.$$Hence, $r \in S$ is equivalent to $(s-r)^2 \ge \frac{4p}{r}$ and $(s-r) + \sqrt{(s-r)^2 - \frac{4p}{r}} \le 2r.$ We clearly have that $S \subseteq [\frac{s}{3}, s).$ Let $T$ be the set of positive reals $r$ in $[\frac{s}{3}, s)$ for which $(s-r)^2 \ge \frac{4p}{r} \Leftrightarrow r(s-r)^2 \ge 4p$. Since $r(s-r)^2$ is decreasing on $[\frac{s}{3}, s),$ it's clear that $T$ is an interval of the form $[\frac{s}{3}, t]$ for some $t \in [\frac{s}{3}, s)$. Now, the second condition is equivalent to: $$\sqrt{(s-r)^2 - \frac{4p}{r}} \le 3r-s \Leftrightarrow (s-r)^2 - \frac{4p}{r} \le (s-3r)^2,$$ where we used that $r \in [\frac{s}{3}, s)$. Upon expanding, the above is equivalent to $p \ge r^2(s-2r)$, so the subset of $T$ satisfying this is clearly an interval since $r^2(s-2r)$ is decreasing on $[\frac{s}{3}, s).$ Hence, the claim is proven. $\blacksquare$ Let $S$ be the set of possible values attained by $\max(a, b, c)$ as $a, b, c \in \mathbb{R}^{+}$ vary while satisfying $abc =p, a+b+c = s.$ $\mathbf{Lemma}$ $\mathbf{2.}$ As $r$ varies in $S$, define $f(r)$ as the value of $ab + bc + ca$ when $a = r$, and $a+b+c = s, abc = p$. Then $r$ is nonincreasing on $S$. ProofNote that $ab+bc+ca = r(b+c) + bc = r(s-r) + \frac{p}{r}.$ Hence, since $S$ is an interval, it only suffices to show that $s - 2r - \frac{p}{r^2} = f'(r) \le 0.$ To show this, we require that $r + r + \frac{p}{r^2} \ge s.$ However, since $b, c \in [\frac{p}{r^2}, r]$ and $bc = r \cdot \frac{p}{r^2},$ it's clear that $b + c \le r + \frac{p}{r^2}$ (say, by factoring the above), and so hence $r + r + \frac{p}{r^2} \ge r + b + c = s,$ as desired. $\blacksquare$ Let's write $d_i = a^i + b^i + c^i$ and $w_i = x^i + y^i + z^i$ for all $ i\in \mathbb{Z}.$ Set $t_i = d_i - w_i$. Observe that: $$d_n = (a+b+c)d_{n-1} - (ab+bc+ca)d_{n-2} + abcd_{n-3} \ge (x+y+z)d_{n-1} - (xy+yz+zx)d_{n-2} + xyzd_{n-3},$$where we used Lemma $2$. From this and a similar expression for $w_n$, we obtain that $t_n \ge (x+y+z) t_{n-1} - (xy+yz+zx)t_{n-2} + xyz t_{n-3}.$ $(1)$ Notice that $t_0 = t_1 = 0$ and $t_2 = 2(xy+yz+zx-ab-bc-ca) \ge 0$ by Lemma $2$. We claim that these values and the inequality above are enough to imply that $t_n \ge 0, \forall n\in \mathbb{N}.$ Define the sequence $\{h_n\}$ by $h_0 = h_1 = 0$, $h_2 = 1$, and $h_{n+3} = (x+y+z)h_{n+2} - (xy+yz+zx)h_{n+1} + xyzh_n$ for $n \ge 0.$ $\mathbf{Claim.}$ $h_i \ge 0, \forall i \in \mathbb{N}.$ ProofSuppose that $x \ge y \ge z$. Let $P_i(x, y, z)$ denote the polynomial $\frac{x^i(y-z) + y^i(z-x) + z^i(x-y)}{(x-y)(y-z)(x-z)}$ (note that the denominator divides the numerator since $x = y, y=z, z=x$ would make the numerator equal to zero). Then, it's easy to see that $h_i = P_i(x, y, z).$ It therefore suffices to prove that $P_i(x, y, z) \ge 0$ for $i \in \mathbb{N}$. Since $P_i$ is clearly continuous in $x, y, z$, we only need to show that $P_i(x, y, z) \ge 0$ when $x>y>z$, since if two of them were equal we can just take the limit as one approaches the other. In that case, it suffices to prove that $x^i(y-z) + y^i(z-x) + z^i(x-y) \ge 0$ when $x > y > z$. In fact, we will show that $x^i(y-z) + y^i(z-x) + z^i(x-y) \ge 0$ as long as $x \ge y \ge z$, which clearly implies that it's true for $x>y>z.$ Let's fix $y, z$. Now, notice that the derivative of the expression with respect to $x$ is $ix^{i-1}(y-z) -y^i + z^i = (y-z) (ix^{i-1} - y^{i-1} - y^{i-2}z - \cdots - yz^{i-2} - z^{i-1}) \ge 0$, where we used the fact that both multiplicands are $\ge 0$. Since we've shown that it's nondecreasing in $x$, and that $x \ge y$, it suffices to verify it when $x = y$. However, it's clear that equality holds in the $x = y$ case, and so the claim is proven. $\blacksquare$ With the claim, we are now ready to show that $t_n \ge 0$ for all $n \in \mathbb{N}.$ We will show by induction on $n+2c$ that: $$t_{n+c} \ge h_{c+2} t_n + (h_{c+3} - (x+y+z)h_{c+2})t_{n-1} + xyzh_{c+1}t_{n-2}$$ for all $n \ge 2, c \ge 1.$ Call the above equation $(2)$. When $c = 1$, it is simply given in $(1)$. Assume it holds for $n+2c \le k$. When $n+2c = k+1,$ we wish to show that: $$t_{n+c} \ge h_{c+2} t_n + (h_{c+3} - (x+y+z)h_{c+2})t_{n-1} + xyzh_{c+1}t_{n-2}.$$Assume that $c > 1$, since otherwise we're done. By the inductive hypothesis, since $(n+1) + 2(c-1) \le k$, we know that: $$t_{n+c} \ge h_{c+1}t_{n+1} + (h_{c+2} - (x+y+z)h_{c+1})t_n + xyzh_ct_{n-1}.$$ However, since $h_{c+1} \ge 0$ by the claim, we have that: $$RHS \ge^{(1)} h_{c+1} ((x+y+z)t_n - (xy+yz+zx)t_{n-1} + xyzt_{n-2}) + (h_{c+2} - (x+y+z)h_{c+1})t_n + xyzh_ct_{n-1} = $$$$h_{c+2}t_n + (xyzh_c - (xy+yz+zx)h_{c+1})t_{n-1} + xyzh_{c+1} t_{n-2}=$$$$h_{c+2}t_n + (h_{c+3} - (x+y+z)h_{c+2})t_{n-1} + xyzh_{c+1}t_{n-2},$$ which is exactly what we wanted. Therefore, the induction is complete, and $(2)$ is proven. By letting $n = 2$, we obtain that for all $c \ge 1$ $t_{c+2} \ge h_{c+2}t_2 \ge 0$, where we used that $h_{c+2}, t_2 \ge 0.$ Therefore, we've shown that $t_{n} \ge 0$ for all $n \ge 3$, and as $t_0, t_1, t_2 \ge 0$ are known, we've shown that $t_n \ge 0$ for all $n \in \mathbb{N}$, which clearly solves the problem. In conclusion, the answer is all $n \ge 0$. $\square$

Any motivation for counter example and proof?

mela_20-15

15.05.2019 17:00

answer: Every $n\ge 0$ works. For $n<0$ , $$(a, b, c) = (2, 3, 4), (x, y, z) = (3.5, \frac{5.5 + \sqrt{5.5^2 - 4*48/7}}{2}, \frac{5.5 - \sqrt{5.5^2 - 4*48/7}}{2})$$Is a counterexample @thanksabove Now $n\ge 0$ . lemma1: If $max(a,b,c)\ge max(x,y,z)$ then $ab+bc+ca\le xy+yz+zx$ It follows from the fact that $(a-x)(a-y)(a-z)\ge (a-a)(a-b)(a-c)$ , $a(xy+yz+zx)\ge a(ab+bc+ca)$ This would be obvious anyway later on the solution, it also proves case $n=2$. We are going to prove the following, using calculus: Fix $a+b+c=s$ , $abc=1$ and denote $ab+bc+ca=p$. $a^n+b^n+c^n$ decreases as $p$ increases. We investigate function $$f(x)=x^2-sx-\frac{1}{x}$$for $x>0$. Note that $f(x)+p=\frac{(x-a)(x-b)(x-c)}{x}$. $f'(x)=2x+\frac{1}{x^2}-s$. As $s>3$ , $f'(1) <0$ figure out that $f'(r_1)=f'(r_2)=0$ for which $r_1<1<r_2<s $ , $f(r_1),f(r_2)<0$. and at this point we come to describe the graph of $f$. $f(x)$ tends to $- \infty$ as $x$ tends to $0$ $f(x)$ increases in $(0,r_1)$ , decreases in $(r_1,r_2)$ , increases in $(r_2,+\infty)$ The gragh of $f$ is under $y=0$ for $x\in(0,s)$. Given the form of $f$ the line $y=-p$ intersects $y=f(x)$ in three points $c<r_1<b<r_2<a$ as long as $-p\in (f(r_2),f(r_1))$ Now we investigate $$g(x)=f(x^{1/n})=x^{2/n}-sx^{1/n}-x^{-1/n}$$. $g$ has the same increasing,decreasing,increasing form and line $y=-p$ intersects $g$ in three different points $c^n<r^n_1<b^n<r^n_2<a^n$ as long as $-p$ is between $g(r^n_2)$ and $g(r^n_1)$. Denote $a(p)=a^n=g^{-1}(-p)$ in interval $(r^n_2,+\infty)$. $g^{-1}$ is differentiable in this interval so $\frac{da(p)}{dp}=\frac{dg^{-1}(-p)}{dp}$ $a'(p)=\frac{-1}{g'(a^n)}$ , $g'(x)=\frac{1}{xn}(2x^{2/n}+x^{-1/n}-sx^{1/n})$ , $g'(a^n)=\frac{1}{na^n}(2a^2+1/a-(a+b+c)a)=\frac{1}{na^n}(a^2+bc-ab-ac)=\frac{1}{na^n}(a-b)(a-c)$. We actually need $a'(p)+b'(p)+c'(p)<0$ (define $b(p),c(p)$ in a similar way) which reduces to $$\frac{a^n}{(a-b)(a-c)}+\frac{b^n}{(b-a)(b-c)}+\frac{c^n}{(c-a)(c-b)}>0$$as $c<b<a$ , $(a-b)(b-c)(a-c)>0$ multiply $$a^n(b-c)+b^n(c-a)+c^n(a-b)>0$$$$(a^n-b^n)(b-c)>(b^n-c^n)(a-b)$$Which is obvious , thus $a(p)+b(p)+c(p)=a^n+b^n+c^n$ decreases as $p$ increases.

dragonheart6

17.05.2019 12:56

If letting $n$ be real number, it changes $\cdots\cdots$ When $0 < n < 1$, it does not work. For example, $n= \frac{1}{2}$, $a = 6, b=3, c=2$ and $x = 5, y = 3 + \frac{3}{\sqrt{5}}, z= 3 - \frac{3}{\sqrt{5}}$. However, $\sqrt{a}+\sqrt{b}+\sqrt{c} < \sqrt{x} + \sqrt{y} + \sqrt{z}$.

dragonheart6

17.05.2019 15:12

It is similar to my old post. https://artofproblemsolving.com/community/u473756h1833674p12287396 Let $a, b, c, u, v, w$ be non-negative real numbers satisfying $$a+ b+ c = u+v+w, \quad ab + bc + ca = uv + vw + wu, \quad abc \ge uvw.$$Let $f(u)$ be a function with $f'''(u)\ge 0, \ \forall u > 0$. Prove that $f(a)+f(b)+f(c) \ge f(u) + f(v) + f(w)$.

mela_20-15

17.05.2019 23:47

Quote: If letting $n$ be real number, it changes $\cdots\cdots$ When $0 < n < 1$, it does not work. For example, $n= \frac{1}{2}$, $a = 6, b=3, c=2$ and $x = 5, y = 3 + \frac{3}{\sqrt{5}}, z= 3 - \frac{3}{\sqrt{5}}$. However, $\sqrt{a}+\sqrt{b}+\sqrt{c} < \sqrt{x} + \sqrt{y} + \sqrt{z}$.] My solution proves the inequality all real $n\ge2$ . I also prove the fact $h(a)+h(b)+h(c)\ge h(x)+h(y)+h(z)$ for three times differentiable $h:[0,+\infty )\to [0,+\infty )$ ,. and $h"(x),h'''(x)>0$. Modify the above arguments so, we basically calculate (I'm silly the definition of $g$ is unnecessery)$$\frac{\partial a}{\partial p}$$but $a=f^{-1}(-p)$ so $$\frac{df^{-1}(-p)}{dp}=-\frac{1}{f'(a)}=-\frac{1}{2a+1/a^2-s}=-\frac{a}{(a-b)(a-c)}$$$$\frac{\partial h(a)}{\partial p}=-\frac{ah'(a)}{(a-b)(a-c)}$$So we have to prove that $(a>b>c)$ $$\frac{ah'(a)}{(a-b)(a-c)}+\frac{bh'(b)}{(b-c)(b-a)}+\frac{ch'(c)}{(c-a)(c-b)}>0$$$$ah'(a)(b-c)+bh'(b)(c-a)+ch'(c)(a-b) >0$$$$\frac{ah'(a)-bh'(b)}{a-b}>\frac{bh'(b)-ch'(c)}{b-c}$$which is trivial by mean value theorem . So here is your problem : Function $f:(0,+\infty )\to (0,+\infty ) $ is three times differentiable and $f''(x),f'''(x)\ge 0$. Positive reals $a,b,c,x,y,z $ are such that $max(a,b,c)\ge max(x,y,z)$ and $a+b+c=x+y+z$ , $abc=xyz$ . Prove that $f(a)+f(b)+f(c)\ge f(x)+f(y)+f(z)$.

dragonheart6

18.05.2019 03:38

Dear mela_20-15, nice solution. Yesterday, I used my method in my old post to get the condition $f''(x) > 0$. https://artofproblemsolving.com/community/c6h1833674p12287396 The main part of my solution is the following: (the idea is similar to the above link) Lemma 1: $F'(q) \le 0$ for $q \in (q_1, q_2)$. Proof of Lemma 1: When $q \in (q_1, q_2)$, $t^3-pt^2+qt-r=0$ has three distinct positive real roots $u, v, w$. WLOG, assume $u > v > w$. Using implicit differentiation for the system of equations $p=u+v+w, \ q = uv+vw+wu, \ r = uvw$, we have $\frac{\partial u}{\partial q} = \frac{-u}{(u-v)(u-w)}, \ \frac{\partial v}{\partial q} = \frac{-v}{(v-u)(v-w)}, \ \frac{\partial w}{\partial q} = \frac{-w}{(w-u)(w-v)}$. Thus, we have $$F'(q) = f'(u)\frac{\partial u}{\partial q} + f'(v) \frac{\partial v}{\partial q} + f'(w) \frac{\partial w}{\partial q} = \frac{-u f'(u)}{(u-v)(u-w)} + \frac{-vf'(v)}{(v-u)(v-w)} + \frac{-wf'(w)}{(w-u)(w-v)}.$$Since $f'''(s)\ge 0, \forall s> 0$, $f'(s)$ is convex on $(0, \infty)$ and hence $\frac{f'(u)-f'(v)}{u-v}\ge \frac{f'(u)-f'(w)}{u-w}$ which results in $f'(u)(v-w) \ge f'(v)(u-w) - f'(w)(u-v)$. Thus, we have $$F'(q) \le \frac{-u (f'(v)(u-w) - f'(w)(u-v)) }{(v-w)(u-v)(u-w)} + \frac{-vf'(v)}{(v-u)(v-w)} + \frac{-wf'(w)}{(w-u)(w-v)} = \frac{-uf'(v)}{(u-v)(v-w)} + \frac{uf'(w)}{(u-w)(v-w)} + \frac{-vf'(v)}{(v-u)(v-w)} + \frac{-wf'(w)}{(w-u)(w-v)} = \frac{-(f'(v)-f'(w))}{v-w}.$$ If $f''(s) > 0$, we have $F'(q) \le 0$.

mela_20-15

18.05.2019 12:52

Yeah it is actually really simple, You need $f'(x_1)+x_1f''(x_1)>f'(x_2)+x_2f''(x_2)$ for$ a>x_1>b>x_2>c$ (from the mean value) So you need the derivative $2f''(x)+xf'''(x)>0$.

K_Mirrorheart

23.05.2019 14:53

Actually during the test, the problem only said: For positive real numbers $a,b,c,x,y,z$, with $max(a,b,c,x,y,z)=a$ , $a+b+c=x+y+z$ and $abc=xyz$, determine all integer $n$ for which the inequality $$a^n+b^n+c^n \ge x^n+y^n+z^n$$holds. This meant that we needed to prove for all $a,b,c,x,y,z$ and $n< 0$, the inequality doesn't hold, rather than giving a counterexample.

Hermitianism

10.06.2019 08:02

im considering whether this problem is related to vasc 's EV theorem to a lesser or greater degree.... (and i believe this problem is presumably dut to hjj)

TheUltimate123

03.11.2020 05:05

Solved with nukelauncher and Th3Numb3rThr33. The answer is $n\ge0$; to verify $n<0$ fail, take $(a,b,c)=(16,3,2)$, $(x,y,z)=(12,8,1)$, where \[\frac1{2^{-n}}+\frac1{3^{-n}}+\frac1{16^{-n}}<1<1+\frac1{8^{-n}}+\frac1{12^{-n}}.\] To prove $n\ge0$ work, we first prove this lemma: Lemma: In $X^3-sX^2-tX-p=0$ with positive real roots $a$, $b$, $c$, if we express $a^n+b^n+c^n$ as a polynomial $P(s,t,p)$, then all terms have positive coefficient. Proof. Easy induction using Newton sums: if $U_n=a^n+b^n+c^n$, then \[U_n=sU_{n-1}+tU_{n-2}+pU_{n-3}.\]$\blacksquare$ Now let $a$, $b$, $c$ be the roots of $X^3-sX^2-t_1X-p=0$ and let $x$, $y$, $z$ be the roots of $X^3-sX^2-t_2X-p=0$. I contend: Claim: $t_1\ge t_2$. Proof. Suppose $t_1<t_2$. Then \[a^3-sa^2-t_2a-p<a^3-sa^2-t_1a-p=0.\]By the intermediate value theorem, one of the roots of $X^3-sX^2-t_2X-p$ is greater than $a$, absurd. $\blacksquare$ Now the desired conclusion follows by combining the claim and the lemma: we have \[a^n+b^n+c^n=P(s,t_1,p)\ge P(s,t_2,p)=x^n+y^n+z^n.\]

csav10

03.11.2020 13:43

Very nice solution, TheUltimate123. Note that Vasc has posted in 1989, in the Romanian journal Gazeta Matematica A, no. 3, the following similar theorem: If $f$ is a three differentiable function on I and $$a_1+a_2+a_3=c_1+c_2+c_3,\ \ \ \ \ \ \ \ a_1^2+a_2^2+a_3^2=c_1^2+c_2^2+c_3^2,$$where $a_i,b_i\in$ I , then there is $x\in$ I such that $$f(a_1)+f(a_2)+f(a_3)-f(c_1)-f(c_2)-f(c_3)=\frac 1{2}(a_1a_2a_3-c_1c_2c_3)f'''(x).$$This means that if $f'''\ge 0$ on I, then $f(a_1)+f(a_2)+f(a_3)-f(c_1)-f(c_2)-f(c_3)\ $ has the same sign as $a_1a_2a_3-c_1c_2c_3$ (which has the sign of $\max\{a_1,a_2,a_3\}-\max\{c_1,c_2,c_3\}).$