let $a_1,a_2,...,a_n$,$b_1,b_2,...,b_n$,$c_1,c_2,...,c_n$ be real numbers. prove that $$ \sum_{cyc}{ \sqrt{\sum_{i \in \{1,...,n\} }{ (3a_i-b_i-c_i)^2}}} \ge \sum_{cyc}{\sqrt{\sum_{i \in \{1,2,...,n\}}{a_i^2}}}$$
Problem
Source: Iranian Third Round 2020 Algebra exam Problem2
Tags: inequalities
21.11.2020 14:28
Kinda liked this one By cauchy we have : $$ \sqrt{\sum(3a_i -b_i -c_i)^2} \geqslant \frac{\sum a_i (3a_i -b_i -c_i)}{\sqrt{\sum a_i ^2}}=3\sqrt{\sum a_i ^2} -\frac{\sum a_i (b_i + c_i)}{\sqrt{\sum a_i ^2}}$$Which using cauchy again gives us $$\geqslant 3\sqrt{\sum a_i ^2} -\sqrt{\sum b_i ^2} -\sqrt{\sum c_i ^2}$$Combining gives us the problem
22.11.2020 12:27
) Now notice that if for some $k$ we have $a_k\ge b_k\ge c_k$ then the sequence $X_k=3a_k-b_k-c_k,Y_k=3b_k-a_k-c_k,Z_k=3c_k-a_k-b_k$ Majorizes The sequence $a_k,b_k,c_k$. Simply note that $X_k=a_k+(a_k-b_k)+(a_k-c_k)\ge a_k$ and $X_k+Y_k=a_k+b_k+(a_k+b_k-2c_k)\ge a_k+b_k$ and obviously $X_k+Y_k+Z_k=a_k+b_k+c_k$. Thus by Karamata's Inequality and the fact that $(X_k,Y_k,Z_k)\succ(a_k,b_k,c_k)$ we'll have: $$f(X_k)+f(Y_k)+f(Z_k)\ge f(a_k)+f(b_k)+f(c_k)$$ Now by considering $f(X_1,X_2,\dots,X_n)=\sqrt{X_1^2+X_2^2+\dots+X_n^2}$ and applying the above for each $k$ in turn, we receive $$f(X_1,X_2,\dots,X_n)+f(Y_1,Y_2,\dots,Y_n)+f(Z_1,Z_2,\dots,Z_n)\ge f(a_1,X_2,\dots,X_n)+f(b_1,Y_2,\dots,Y_n)+f(c_1,Z_2,\dots,Z_n)\ge$$$$\cdot$$$$\cdot$$$$\cdot$$$$\ge f(a_1,a_2,\dots,a_n)+f(b_1,b_2,\dots,b_n)+f(c_1,c_2,\dots,c_n)$$Which is the result we want!
28.11.2020 07:22
H.M-Deadline wrote: Thus by Karamata's Inequality and the fact that $(X_k,Y_k,Z_k)\succ(a_k,b_k,c_k)$ we'll have: $$f(X_k)+f(Y_k)+f(Z_k)\ge f(a_k)+f(b_k)+f(c_k)$$ Now by considering $f(X_1,X_2,\dots,X_n)=\sqrt{X_1^2+X_2^2+\dots+X_n^2}$ and applying the above for each $k$ in turn, we receive $$f(X_1,X_2,\dots,X_n)+f(Y_1,Y_2,\dots,Y_n)+f(Z_1,Z_2,\dots,Z_n)\ge f(a_1,X_2,\dots,X_n)+f(b_1,Y_2,\dots,Y_n)+f(c_1,Z_2,\dots,Z_n)\ge$$$$\cdot$$$$\cdot$$$$\cdot$$$$\ge f(a_1,a_2,\dots,a_n)+f(b_1,b_2,\dots,b_n)+f(c_1,c_2,\dots,c_n)$$Which is the result we want! Now do correct me if I’m wrong (as I’m not very well versed in karamata) but when we use karamata inequality on $f(X_1,X_2,\dots,X_n)+f(Y_1,Y_2,\dots,Y_n)+f(Z_1,Z_2,\dots,Z_n)\ge f(a_1,X_2,\dots,X_n)+f(b_1,Y_2,\dots,Y_n)+f(c_1,Z_2,\dots,Z_n)$ Then $g(X_1)=f(X_1,X_2,\dots,X_n)=\sqrt{X_1 ^2 + a}$ but now if we change for example $X_2$ the function doesn’t stay the same because “$a$” differs. so $g(X_1) ,g(Y_1) ,g(Z_1)$ (which are the functions we used karamata on) are not necessarily the same function as their “$a$” can differ. Can we still use karamata on different functions?
28.11.2020 11:14
TheBarioBario wrote: H.M-Deadline wrote: Thus by Karamata's Inequality and the fact that $(X_k,Y_k,Z_k)\succ(a_k,b_k,c_k)$ we'll have: $$f(X_k)+f(Y_k)+f(Z_k)\ge f(a_k)+f(b_k)+f(c_k)$$ Now by considering $f(X_1,X_2,\dots,X_n)=\sqrt{X_1^2+X_2^2+\dots+X_n^2}$ and applying the above for each $k$ in turn, we receive $$f(X_1,X_2,\dots,X_n)+f(Y_1,Y_2,\dots,Y_n)+f(Z_1,Z_2,\dots,Z_n)\ge f(a_1,X_2,\dots,X_n)+f(b_1,Y_2,\dots,Y_n)+f(c_1,Z_2,\dots,Z_n)\ge$$$$\cdot$$$$\cdot$$$$\cdot$$$$\ge f(a_1,a_2,\dots,a_n)+f(b_1,b_2,\dots,b_n)+f(c_1,c_2,\dots,c_n)$$Which is the result we want! Now do correct me if I’m wrong (as I’m not very well versed in karamata) but when we use karamata inequality on $f(X_1,X_2,\dots,X_n)+f(Y_1,Y_2,\dots,Y_n)+f(Z_1,Z_2,\dots,Z_n)\ge f(a_1,X_2,\dots,X_n)+f(b_1,Y_2,\dots,Y_n)+f(c_1,Z_2,\dots,Z_n)$ Then $g(X_1)=f(X_1,X_2,\dots,X_n)=\sqrt{X_1 ^2 + a}$ but now if we change for example $X_2$ the function doesn’t stay the same because “$a$” differs. so $g(X_1) ,g(Y_1) ,g(Z_1)$ (which are the functions we used karamata on) are not necessarily the same function as their “$a$” can differ. Can we still use karamata on different functions? Actually this is exactly why I stated the function is convex for any nonnegative $a$ since Karamata's inequality has been generalized as Schur/Hardy-Littlewood-Polya/Karamata inequality for symmetric convex functions of several variables, You can find more on this in the books $Analytic\ Inequalities$ in it's second edition as well as $Inequalities:\ Theory\ of\ Majorization\ and\ its\ applications$, or you can look through Google scholarship as there are papers on it or use it for the sake of proving something else!
07.12.2020 22:51
Hi Does anyone have any idea how to continue this solution? frame|center|772px|aops
22.05.2022 16:59
Claim: $\sqrt{{a_1}^2+\cdots +{a_n}^2} +\sqrt{{b_1}^2+\cdots +{b_n}^2} +\sqrt{{c_1}^2+\cdots +{c_n}^2} \geq \sqrt{(a_1+b_1+c_1)^2+\cdots +(a_n+b_n+c_n)^2}$ By the claim we have: $\sum{ \sqrt{\sum_{i \in \{1,...,n\} }{ (3a_i-b_i-c_i)^2}}} + \sum{ \sqrt{\sum_{i \in \{1,...,n\} }{ (3b_i-a_i-c_i)^2}}} \geq 2\sqrt{\sum_{i \in \{1,...,n\} }{ (a_i+b_i-c_i)^2}}$ $\sum{ \sqrt{\sum_{i \in \{1,...,n\} }{ (3b_i-a_i-c_i)^2}}} + \sum{ \sqrt{\sum_{i \in \{1,...,n\} }{ (3c_i-a_i-b_i)^2}}} \geq 2\sqrt{\sum_{i \in \{1,...,n\} }{ (b_i+c_i-a_i)^2}}$ $\sum{ \sqrt{\sum_{i \in \{1,...,n\} }{ (3a_i-b_i-c_i)^2}}} + \sum{ \sqrt{\sum_{i \in \{1,...,n\} }{ (3c_i-a_i-b_i)^2}}} \geq 2\sqrt{\sum_{i \in \{1,...,n\} }{ (a_i+c_i-b_i)^2}}$ Thus we have:$$ \sum_{cyc}{ \sqrt{\sum_{i \in \{1,...,n\} }{ (3a_i-b_i-c_i)^2}}} \ge \sum_{cyc}{ \sqrt{\sum_{i \in \{1,...,n\} }{ (a_i+b_i-c_i)^2}}}$$again by the claim we have: $\sqrt{\sum_{i \in \{1,...,n\} }{ (a_i+b_i-c_i)^2}}+\sqrt{\sum_{i \in \{1,...,n\} }{ (a_i+c_i-b_i)^2}} \ge 2\sum{\sqrt{\sum_{i \in \{1,2,...,n\}}{a_i^2}}}$ $\sqrt{\sum_{i \in \{1,...,n\} }{ (b_i+c_i-a_i)^2}}+\sqrt{\sum_{i \in \{1,...,n\} }{ (b_i+a_i-c_i)^2}} \ge 2\sum{\sqrt{\sum_{i \in \{1,2,...,n\}}{b_i^2}}}$ $\sqrt{\sum_{i \in \{1,...,n\} }{ (b_i+c_i-a_i)^2}}+\sqrt{\sum_{i \in \{1,...,n\} }{ (a_i+c_i-b_i)^2}} \ge 2\sum{\sqrt{\sum_{i \in \{1,2,...,n\}}{c_i^2}}}$ Thus we have:$$\sum_{cyc}{ \sqrt{\sum_{i \in \{1,...,n\} }{ (a_i+b_i-c_i)^2}}} \ge \sum_{cyc}{\sqrt{\sum_{i \in \{1,2,...,n\}}{a_i^2}}}$$so we are done.
26.06.2022 09:40
Mahdi.sh wrote: Claim: $\sqrt{{a_1}^2+\cdots +{a_n}^2} +\sqrt{{b_1}^2+\cdots +{b_n}^2} +\sqrt{{c_1}^2+\cdots +{c_n}^2} \geq \sqrt{(a_1+b_1+c_1)^2+\cdots +(a_n+b_n+c_n)^2}$ By the claim we have: $\sum{ \sqrt{\sum_{i \in \{1,...,n\} }{ (3a_i-b_i-c_i)^2}}} + \sum{ \sqrt{\sum_{i \in \{1,...,n\} }{ (3b_i-a_i-c_i)^2}}} \geq 2\sqrt{\sum_{i \in \{1,...,n\} }{ (a_i+b_i-c_i)^2}}$ $\sum{ \sqrt{\sum_{i \in \{1,...,n\} }{ (3b_i-a_i-c_i)^2}}} + \sum{ \sqrt{\sum_{i \in \{1,...,n\} }{ (3c_i-a_i-b_i)^2}}} \geq 2\sqrt{\sum_{i \in \{1,...,n\} }{ (b_i+c_i-a_i)^2}}$ $\sum{ \sqrt{\sum_{i \in \{1,...,n\} }{ (3a_i-b_i-c_i)^2}}} + \sum{ \sqrt{\sum_{i \in \{1,...,n\} }{ (3c_i-a_i-b_i)^2}}} \geq 2\sqrt{\sum_{i \in \{1,...,n\} }{ (a_i+c_i-b_i)^2}}$ Thus we have:$$ \sum_{cyc}{ \sqrt{\sum_{i \in \{1,...,n\} }{ (3a_i-b_i-c_i)^2}}} \ge \sum_{cyc}{ \sqrt{\sum_{i \in \{1,...,n\} }{ (a_i+b_i-c_i)^2}}}$$again by the claim we have: $\sqrt{\sum_{i \in \{1,...,n\} }{ (a_i+b_i-c_i)^2}}+\sqrt{\sum_{i \in \{1,...,n\} }{ (a_i+c_i-b_i)^2}} \ge 2\sum{\sqrt{\sum_{i \in \{1,2,...,n\}}{a_i^2}}}$ $\sqrt{\sum_{i \in \{1,...,n\} }{ (b_i+c_i-a_i)^2}}+\sqrt{\sum_{i \in \{1,...,n\} }{ (b_i+a_i-c_i)^2}} \ge 2\sum{\sqrt{\sum_{i \in \{1,2,...,n\}}{b_i^2}}}$ $\sqrt{\sum_{i \in \{1,...,n\} }{ (b_i+c_i-a_i)^2}}+\sqrt{\sum_{i \in \{1,...,n\} }{ (a_i+c_i-b_i)^2}} \ge 2\sum{\sqrt{\sum_{i \in \{1,2,...,n\}}{c_i^2}}}$ Thus we have:$$\sum_{cyc}{ \sqrt{\sum_{i \in \{1,...,n\} }{ (a_i+b_i-c_i)^2}}} \ge \sum_{cyc}{\sqrt{\sum_{i \in \{1,2,...,n\}}{a_i^2}}}$$so we are done. How do you claim your first line?
29.07.2022 20:46
According to Minkowski's inequality, we can say : $\left ( \sqrt{\sum_{i = 1}^{n}(3a_{i}-b_{i}-c_{i})} \right )+\left ( \sqrt{\sum_{i=1}^{n}}b^2_{i} \right )+\left ( \sqrt{\sum_{i=1}^{n}}c^2_{i} \right ) \geq \sqrt{\sum_{i=1}^{n}(3a_{i})^2}$ If we write and add the same inequality for other variables, the verdict is easily obtained !
29.07.2022 21:41
Let $\mathbf{a}=(a_1,\cdots,a_n)$, $\mathbf{b}=(b_1,\cdots,b_n)$ and $\mathbf{c}=(c_1,\cdots,c_n)$ be vectors in $\mathbb{R}^n$. First, note that the Triangle Inequality tells us that, for any two vectors $\mathbf{x},\mathbf{y} \in \mathbb{R}^n$, $\Vert \mathbf{x} \Vert + \Vert \mathbf{y} \Vert \geq \Vert \mathbf{x}+\mathbf{y} \Vert$ (this is equivalent to the usage of Minkowski's Inequality in other solutions). Using vector notation, we want to prove that $\Vert 3\mathbf{a} - \mathbf{b} - \mathbf{c} \Vert + \Vert 3\mathbf{b} - \mathbf{c} - \mathbf{a} \Vert + \Vert 3\mathbf{c} - \mathbf{a} - \mathbf{b} \Vert \geq \Vert \mathbf{a} \Vert + \Vert \mathbf{b} \Vert + \Vert \mathbf{c} \Vert$. The Triangle Inequality tells us that $\Vert 3\mathbf{a} - \mathbf{b} - \mathbf{c} \Vert + \Vert 3\mathbf{b} - \mathbf{c} - \mathbf{a} \Vert \geq \Vert 2\mathbf{a} + 2\mathbf{b} - 2\mathbf{c} \Vert = 2\Vert \mathbf{a} + \mathbf{b} - \mathbf{c} \Vert$; summing cyclically over $\mathbf{a},\mathbf{b} ,\mathbf{c}$ and dividing by $2$ tells us that $\Vert 3\mathbf{a} - \mathbf{b} - \mathbf{c} \Vert + \Vert 3\mathbf{b} - \mathbf{c} - \mathbf{a} \Vert + \Vert 3\mathbf{c} - \mathbf{a} - \mathbf{b} \Vert \geq \Vert \mathbf{a} + \mathbf{b} - \mathbf{c} \Vert + \Vert \mathbf{b} + \mathbf{c} - \mathbf{a} \Vert + \Vert \mathbf{c} + \mathbf{a} - \mathbf{b} \Vert$. Once again applying the Triangle Inequality, $\Vert \mathbf{a} + \mathbf{b} - \mathbf{c} \Vert + \Vert \mathbf{b} + \mathbf{c} - \mathbf{a} \Vert \geq 2 \Vert \mathbf{b} \Vert$; summing cyclically over $\mathbf{a},\mathbf{b} ,\mathbf{c}$ and dividing by $2$ tells us that $\Vert \mathbf{a} + \mathbf{b} - \mathbf{c} \Vert + \Vert \mathbf{b} + \mathbf{c} - \mathbf{a} \Vert + \Vert \mathbf{c} + \mathbf{a} - \mathbf{b} \Vert \geq \Vert \mathbf{a} \Vert + \Vert \mathbf{b} \Vert + \Vert \mathbf{c} \Vert$, and combining both of the inequalities we found gives us the desired result.
20.03.2024 01:10
Proposed by Mojtaba Zare.