Let $a, b, c$ be nonnegative real numbers. Prove that \[ \frac{a+b+c}{3} - \sqrt[3]{abc} \leq \max\{(\sqrt{a} - \sqrt{b})^2, (\sqrt{b} - \sqrt{c})^2, (\sqrt{c} - \sqrt{a})^2\}. \]
Problem
Source: USA TST 2000
Tags: inequalities
18.11.2005 14:47
http://www.mathlinks.ro/Forum/viewtopic.php?t=12554 http://www.mathlinks.ro/Forum/viewtopic.php?t=14014 darij
18.11.2005 15:55
Actualy, if $A=\{(\sqrt{a} - \sqrt{b})^2, (\sqrt{b} - \sqrt{c})^2, (\sqrt{c} - \sqrt{a})^2\}$ and $a\geq0,b\geq0,c\geq0$ then $\frac{1}{2}minA\leq\frac{a+b+c}{3}-\sqrt[3]{abc}\leq maxA.$
02.06.2006 14:03
What about the genaralisation ? If $a_1,...,a_n\geq 0$ then prove that $\frac{m}{2}\leq\frac{a_1+a_2+.....+a_n}{n}-\sqrt[n]{a_1a_2.....a_n}\leq\frac{(n-1)M}{2}$ where $m=min_{1\leq i<j\leq n}\left\{ (\sqrt{a_i}-\sqrt{a_j})^2\right\}$ and $M=max_{1\leq i<j\leq n}\left\{ (\sqrt{a_i}-\sqrt{a_j})^2\right\}$
14.03.2010 15:22
we should prove that LHS is smaller than the average and the problem is solved. $ a + b + c - 3\sqrt [3]{abc}\le{2(a + b + c) - 2(\sqrt {ab} + \sqrt {bc} + \sqrt {ca})}$ $ a + b + c + 3\sqrt [3]{abc}\ge{2(\sqrt {ab} + \sqrt {bc} + \sqrt {ca})}$ $ a = x^6,b = y^6,c = z^6$ by schur inequality: $ x^6 + y^6 + z^6 + 3x^2y^2z^2\ge{\sum{x^2y^4} + \sum{x^4y^2}}$ and by AM-GM: $ \sum{x^4y^2} + \sum{x^2y^4}\ge{2\sum{x^3y^3}}$ proved!
14.12.2015 23:42
Since $\max\{(\sqrt{a} - \sqrt{b})^2, (\sqrt{b} - \sqrt{c})^2, (\sqrt{c} - \sqrt{a})^2\}\ge \dfrac{(\sqrt{a} - \sqrt{b})^2+(\sqrt{b} - \sqrt{c})^2+ (\sqrt{c} - \sqrt{a})^2}{3}=\dfrac{2}{3}(a+b+c-\sqrt{ab}-\sqrt{bc}-\sqrt{ca})$ so it remains to prove after some algebra $$2(\sqrt{ab}+\sqrt{bc}+\sqrt{ca})\le a+b+c+3\sqrt[3]{abc}$$However, note that $2(\sqrt{ab}+\sqrt{bc}+\sqrt{ca})\le \sum_{sym}\sqrt[3]{a^2b}$ by AM-GM and the problem reduces to Schur.
08.03.2016 00:01
Observe that \[ \max\{(\sqrt{a} - \sqrt{b})^2, (\sqrt{b} - \sqrt{c})^2, (\sqrt{c} - \sqrt{a})^2\} \ge \frac{(\sqrt{a}-\sqrt{b})^2+(\sqrt{b}-\sqrt{c})^2+(\sqrt{c}-\sqrt{a})^2}{3} \]We wish to show that, upon some expanding, that \[ \frac{a+b+c}{3} + \sqrt[3]{abc} \ge \frac{2}{3} (\sqrt{ab}+\sqrt{bc}+\sqrt{ca}) \]Let $a=x^6, b=y^6, c=z^6$. Then by degree three Schur on $x^2, y^2, z^2$ and Muirhead we have \[ x^6+y^6+z^6+3x^2y^2z^2 \ge \sum_{sym} x^4y^2 \ge 2(x^3y^3+y^3z^3+z^3x^3) \]as requested.
19.08.2017 16:07
Here is an excellent solution: Let $a=x^6, b=y^6, c=z^6$, with WLOG $x\geq y\geq z$. Then the inequality turns into \[x^6+y^6+z^6-3x^2y^2z^2\leq 3(z^3-x^3)^2 \implies y^6+6x^3z^3\leq 2(x^6+z^6)+3x^2y^2z^2.\]Let $y=z+m$ and $x=z+m+n$ where $m, n$ are nonnegative numbers. It suffices to prove that \[2(z+m+n)^6+2z^6+3z^2(z+m)^2(z+m+n)^2-(z+m)^6-6z^3(z+m+n)^3\geq 0,\]or \[m^6 + 12 m^5 n + 6 m^5 z + 30 m^4 n^2 + 60 m^4 n z + 18 m^4 z^2 + 40 m^3 n^3 + 120 m^3 n^2 z + 126 m^3 n z^2 + 26 m^3 z^3 + 30 m^2 n^4 + 120 m^2 n^3 z + 183 m^2 n^2 z^2 + 120 m^2 n z^3 + 15 m^2 z^4 + 12 m n^5 + 60 m n^4 z + 120 m n^3 z^2 + 108 m n^2 z^3 + 42 m n z^4 + 2 n^6 + 12 n^5 z + 30 n^4 z^2 + 34 n^3 z^3 + 15 n^2 z^4\geq 0,\]which is trivially true. Equality holds when $m=n=0$, or $a=b=c$.
09.02.2019 02:50
WLOG, $a \le b \le c$. Fix $a,c$, the function $f(b) = \frac{a+b+c}{3} - \sqrt[3]{abc}$ is convex and thus achieves a maximum when $b= a$ or $b=c$. Let $b =a$, then it suffices to prove $2(c/a) -6(c/a)^{1/2} + 3(c/a)^{1/3} +1 \ge 0$. If $b= c$ then it suffices to prove $2(a/c) - 6(a/c)^{1/2} + 3(a/c)^{1/3} +1 \ge 0$. Note that the polynomial $2x^6 + 3x^2 -6x^3 +1 = (x-1)^2(2x^4 + 4 x^3 + 6x^2+2x +1)$ which is nonnegative for all $x>0$. The inequality is proved.
28.08.2019 14:07
MithsApprentice wrote: Let $a, b, c$ be nonnegative real numbers. Prove that \[ \frac{a+b+c}{3} - \sqrt[3]{abc} \leq \max\{(\sqrt{a} - \sqrt{b})^2, (\sqrt{b} - \sqrt{c})^2, (\sqrt{c} - \sqrt{a})^2\}. \] Claim : $x^6+y^6+z^6+3x^2y^2z^2\ge 2(x^3y^3+y^3z^3+z^3x^3)$ Proof : Since $x^2+y^2\ge 2xy$ and By Schur's Inequality we get $$x^6+y^6+z^6+3x^2y^2z^2\ge x^2y^2(x^2+y^2)+y^2z^2(y^2+z^2)+z^2x^2(z^2+x^2)\ge 2(x^3y^3+y^3z^3+z^3x^3)$$so the claim is proven.$\square$ Back to the original problem FTSOC assume \[ \frac{a+b+c}{3} - \sqrt[3]{abc} \geq \max\{(\sqrt{a} - \sqrt{b})^2, (\sqrt{b} - \sqrt{c})^2, (\sqrt{c} - \sqrt{a})^2\}. \].Adding the cyclic variants we get $a+b+c-3\sqrt[3]{abc}\geq (\sqrt{a} - \sqrt{b})^2+(\sqrt{b} - \sqrt{c})^2+(\sqrt{c} - \sqrt{a})^2 \Rightarrow 2(\sqrt{ab}+\sqrt{bc}+\sqrt{ca})\geq a+b+c+3\sqrt[3]{abc}$. Plugging $x=\sqrt[6]{a}$,$y=\sqrt[6]{b}$,$z=\sqrt[6]{c}$ we get $2(x^3y^3+y^3z^3+z^3x^3)\geq x^6+y^6+z^6+3x^2y^2z^2$ a contradiction to our claim so we are done.$\blacksquare$
25.03.2020 01:05
We substitute $a=x^6$, $b=y^6$, $c=z^6$, and WLOG $x\leq y \leq z$. The key idea is to use the following factorization of the LHS: $$\frac{x^6+y^6+z^6}{3}-x^2y^2z^2=\frac{1}{6}(x^2+y^2+z^2)(\sum (x^2-y^2)^2) \leq \frac{1}{6}(x^2+z^2+z^2)(3(x^2-z^2)^2)\leq (x^2+z^2)(x^2-z^2)^2.$$and now, the result follows by AM-GM since $$(x^3-z^3)^2=(x-z)^2\left(\frac{x^2+z^2+(x+z)^2}{2} \right)^2 \geq (x-z)^2(x+z)^2(x^2+z^2) = (x^2-z^2)^2(x^2+y^2)$$ as desired
13.02.2021 14:19
Since $\max\{(\sqrt{a} - \sqrt{b})^2, (\sqrt{b} - \sqrt{c})^2, (\sqrt{c} - \sqrt{a})^2\} \ge \frac{(\sqrt a-\sqrt b)^2+(\sqrt b-\sqrt c)^2+(\sqrt c-\sqrt a)^2}{3}=\frac{2(a+b+c)}{3}-\frac{2(\sqrt{ab}+\sqrt{bc}+\sqrt{ca})}{3}$, we ought to prove that \[a+b+c+3\sqrt[3]{abc} \ge 2(\sqrt{ab}+\sqrt{bc}+\sqrt{ca}).\]By using Schur Inequality, we have \[a+b+c+3\sqrt[3]{abc}\ge (\sqrt[3]{a^2b}+\sqrt[3]{ab^2})+(\sqrt[3]{b^2c}+\sqrt[3]{bc^2})+(\sqrt[3]{c^2a}+\sqrt[3]{ca^2})\ge 2(\sqrt{ab}+\sqrt{bc}+\sqrt{ca})\]with the use of AM-GM at last. $\blacksquare$
04.08.2021 20:49
Suppose otherwise, so $$\frac{a+b+c}{3}-\sqrt[3]{abc}>\max\{(\sqrt{a} - \sqrt{b})^2, (\sqrt{b} - \sqrt{c})^2, (\sqrt{c} - \sqrt{a})^2\} \implies \frac{a+b+c}{3}-\sqrt[3]{abc}>(\sqrt{a}-\sqrt{b})^2$$and all cyclic variations. So summing cyclically: $$a+b+c-3\sqrt[3]{abc}>\sum_{\mathrm{cyc}} (\sqrt{a}-\sqrt{b})^2=2(a+b+c)-2\sum_\mathrm{cyc} \sqrt{ab} \implies a+b+c+3\sqrt[3]{abc} \leq 2\sum_\mathrm{cyc} \sqrt{ab}.$$But by Schur's and Muirhead, as $(\tfrac{2}{3},\tfrac{1}{3},0) \succ (\tfrac{1}{2},\tfrac{1}{2},0)$: $$a+b+c+3\sqrt[3]{abc} \geq \sum_\mathrm{sym} \sqrt[3]{a^2b}\geq 2\sum_\mathrm{cyc} \sqrt{ab},$$which is a contradiction, so the desired inequality is true. $\blacksquare$
09.08.2023 11:39
WLOG $a\geq b\geq c$. Then, this becomes $$\frac{a+b+c}{3}-\sqrt[3]{abc}\leq (\sqrt{a}-\sqrt{c})^2$$$$a+b+c-3\sqrt[3]{abc}=3a+3c-6\sqrt{ac}$$$$2a+2c+3\sqrt[3]{abc}\geq b+6\sqrt{ac}.$$This is a homogenous inequality, so WLOG $ac=1$. Then, this is $$2a+2c+3\sqrt[3]{b}-b\geq 6.$$Note that now, all of the terms involving $b$ are by themselves. This means that we only have to check the "worst case", which is when $3\sqrt[3]{b}-b$ is at the minimum, since this is the only part of the equation that $b$ affects. The derivative of this is $\frac{1}{b^{2/3}}-1$. which means that the function $3\sqrt[3]{b}-b$ increases on $(0,1]$ and decreases on $[1,\infty)$. Thus, given any interval that contains only nonnegatives, the minimum value in that interval has to occur at either the left or right endpoint. Thus, it suffices to show the inequality for the cases $b=a$ and $b=c$. When $b=a$, we have $$a+2c+3\sqrt[3]{a}\geq 6$$$$a+\frac{2}{a}+3\sqrt[3]{a}\geq 6.$$The derivative of the left side is $$\frac{a^2+a^{4/3}-2}{a^2},$$so it will decrease up until $a=1$ and then increase past that, so the minimum value occurs at $a=1$ which gives 6, which shows this case. When $b=c$ we have $$2a+c+3\sqrt[3]{c}\geq 6$$$$\frac{2}{c}+c+3\sqrt[3]{c}\geq 6,$$which is the exact same thing, hence done.
31.08.2023 07:23
Refinement and generalization, here https://artofproblemsolving.com/community/c6t243f6h3117508_httpsartofproblemsolvingcomcommunityc3339224