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

) 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!

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?

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!

Hi Does anyone have any idea how to continue this solution? frame|center|772px|aops

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.

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?

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 !

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.

Proposed by Mojtaba Zare.