Show that for all positive real numbers $a, b, c$, we have 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: 2023 Brazil Ibero TST P3
Tags: inequalities, Inequality, arithmetic mean, geometric mean, max
mihaig
28.08.2023 18:31
YLG_123 wrote: Show that for all positive real numbers $a, b, c$, we have that $$\frac{a+b+c}{3}-\sqrt[3]{abc} \leq \max\{(\sqrt{a}-\sqrt{b})^2, (\sqrt{b}-\sqrt{c})^2, (\sqrt{c}-\sqrt{a})^2\}$$ $$\frac{a+b+c}{3}-\sqrt[3]{abc} \leq \frac23\cdot\max\{(\sqrt{a}-\sqrt{b})^2, (\sqrt{b}-\sqrt{c})^2, (\sqrt{c}-\sqrt{a})^2\}$$(V. Cirtoaje)
MathSaiyan
28.08.2023 20:18
We can just prove that
\[
\frac{a+b+c}{3}-\sqrt[3]{abc} \leq \frac 13\left((\sqrt{a}-\sqrt{b})^2 + (\sqrt{b}-\sqrt{c})^2 + (\sqrt{c}-\sqrt{a})^2\right),
\]which just follows by plugging $a=x^3,b=y^3,c=z^3$ and expanding, which reduces the problem to
\[
2\left((xy)^{\frac 32}+(yz)^{\frac 32}+(zx)^{\frac 32}\right) \le x^3+y^3+z^3+3xyz
\]but by Muirhead (left ineq) and Schur (right ineq):
\[
2\left((xy)^{\frac 32}+(yz)^{\frac 32}+(zx)^{\frac 32}\right) \le x^2y+xy^2+y^2z+yz^2+z^2x+zx^2 \le x^3+y^3+z^3+3xyz
\]so we win.
grupyorum
28.08.2023 20:35
Suppose the assertion is false for some $a,b,c>0$. Let $a=m^6$, $b=n^6$ and $c=k^6$. We find \[ \frac{m^6+n^6+k^6}{3} - m^2n^2k^2>\max\left\{\left(m^3-n^3\right)^2,\left(n^3-k^3\right)^2,\left(k^3-m^3\right)^2\right\}. \]Summing these up, we find \[ 2m^3n^3 + 2n^3k^3+2k^3m^3 > m^6+n^6+k^6 + 3m^2n^2k^2. \]Next, we have by Schur's inequality \[ \sum m^2(m^2-n^2)(m^2-k^2)\ge 0 \Leftrightarrow \sum m^6 + 3m^2n^2k^2 \ge [4,2,0], \]where $[4,2,0]$ is the sum of all six terms of form $m^\alpha n^\beta k^\gamma$ with $\{\alpha,\beta,\gamma\}=\{4,2,0\}$. Hence, we obtain $[3,3,0]>[4,2,0]$, which is absurd by Muirhead's since $[4,2,0]$ majorizes $[3,3,0]$.
IAmTheHazard
28.08.2023 20:53
USA TST 2000/1
mihaig
31.08.2023 03:56
MathSaiyan wrote:
We can just prove that
\[
\frac{a+b+c}{3}-\sqrt[3]{abc} \leq \frac 13\left((\sqrt{a}-\sqrt{b})^2 + (\sqrt{b}-\sqrt{c})^2 + (\sqrt{c}-\sqrt{a})^2\right),
\]which just follows by plugging $a=x^3,b=y^3,c=z^3$ and expanding, which reduces the problem to
\[
2\left((xy)^{\frac 32}+(yz)^{\frac 32}+(zx)^{\frac 32}\right) \le x^3+y^3+z^3+3xyz
\]but by Muirhead (left ineq) and Schur (right ineq):
\[
2\left((xy)^{\frac 32}+(yz)^{\frac 32}+(zx)^{\frac 32}\right) \le x^2y+xy^2+y^2z+yz^2+z^2x+zx^2 \le x^3+y^3+z^3+3xyz
\]so we win.
Funny, but false. Check $a=b=1$ and $c=0.$
mihaig
31.08.2023 04:24
mihaig wrote: YLG_123 wrote: Show that for all positive real numbers $a, b, c$, we have that $$\frac{a+b+c}{3}-\sqrt[3]{abc} \leq \max\{(\sqrt{a}-\sqrt{b})^2, (\sqrt{b}-\sqrt{c})^2, (\sqrt{c}-\sqrt{a})^2\}$$ $$\frac{a+b+c}{3}-\sqrt[3]{abc} \leq \frac23\cdot\max\{(\sqrt{a}-\sqrt{b})^2, (\sqrt{b}-\sqrt{c})^2, (\sqrt{c}-\sqrt{a})^2\}$$(V. Cirtoaje) See, of course, https://artofproblemsolving.com/community/c6t243f6h3117508_httpsartofproblemsolvingcomcommunityc3339224
bin_sherlo
01.11.2024 19:17
Suppose $a\geq b\geq c$. \[f(a,b,c)=3(\sqrt{a}-\sqrt{c})^2+3\sqrt[3]{abc}-a-b-c=2a+2c-b-6\sqrt{ac}+3\sqrt[3]{abc}\]We want to show that $f(a,b,c)\geq 0$. Let $a,b,c$ be the triplet where $f_{min}$ is achieved. If $ac\geq b^2$, then \[f(a,b,c)-f(\frac{ac}{b},b,b)=2a+2c-b-6\sqrt{ac}+3\sqrt[3]{abc}-\frac{2ac}{b}-2b+b+6\sqrt{ac}-3\sqrt[3]{abc}=\frac{(b-c)(2a-b)}{b}\geq 0\]Similarily if $b^2\geq ac$, then $f(b,b,\frac{ac}{b})$. Let $\frac{ac}{b}=t$. \[f(a,b,c)\geq 2t+b-6\sqrt{bt}+3\sqrt[3]{b^2t}=(t+t+b+\sqrt[3]{b^2t}+\sqrt[3]{b^2t}+\sqrt[3]{b^2t})-6\sqrt{bt}\overset{AM-GM}{\geq} 0\]As desired.$\blacksquare$