It's obvious by Holder's inequality. By Holder's inequality

Raja Oktovin wrote: Let $ a$, $ b$, and $ c$ be non-negative real numbers and let $ x$, $ y$, and $ z$ be positive real numbers such that $ a + b + c = x + y + z$. Prove that \[ \dfrac{a^3}{x^2} + \dfrac{b^3}{y^2} + \dfrac{c^3}{z^2} \ge a + b + c.\] Hery Susanto, Malang $ \frac{a^3}{x^2}+x+x \geq 3a$

By Holder inequality ,we have $ (x+y+z)(x+y+z)\left(\frac{a^3}{x^2} + \frac{b^3}{y^2} + \frac{c^3}{z^2}\right) \ge (a+b+c)^3$ $ \frac{a^3}{x^2} + \frac{b^3}{y^2} + \frac{c^3}{z^2}\ge\frac{(a+b+c)^3}{(x+y+z)^2}$ since $ x+y+z=a+b+c$ so $ \frac{a^3}{x^2} + \frac{b^3}{y^2} + \frac{c^3}{z^2}\ge\frac{(a+b+c)^3}{(x+y+z)^2}=a+b+c$ EDITED : Sorry I didn't open your quoting (dgreenb801) before

I think the problem without the condition $ \sum a =\sum x$ was asked in the Belarus Olympiad 2000....Nevertheless, a useful tool for solving tougher inequalities...

This problem is an imediate result of the following inequality: If: $ a_i\geq0,b_i>0,i=1,2,\dots,k,$ and $ k,n\in N,k\geq1,$ then: $ \frac{{a_1}^{n+1}}{b_1^n}+\frac{a_2^{n+1}}{b_2^n}+\cdots+\frac{{a_k}^{n+1}}{b_k^n}\geq\frac{(a_1+\cdots+a_k)^{n+1}}{(b_1+\cdots+b_k)^n}$ You can notice that I've proved this inequality, using my lemma, at http://www.mathlinks.ro/viewtopic.php?t=310721 , as an application of it. Of course, it could also be considered as a result of Holder's inequality. For $ a_1+\cdots+a_k=b_1+\cdots+b_k$ we obtain: $ \frac{{a_1}^{n+1}}{b_1^n}+\frac{a_2^{n+1}}{b_2^n}+\cdots+\frac{{a_k}^{n+1}}{b_k^n} \geq a_1+\cdots+a_k$ For $ k=3, n=2$ we get to the initial problem.

I don't know what Holder ineq is,but I can second-kill it by using Carlson's ineq: by Carlson's ineq $(\sum\frac{a^3}{x^2})(\sum x)(\sum x)\ge (\sum a)^3$ what follows is trivial.

marin.bancos wrote: This problem is an imediate result of the following inequality: If: $ a_i\geq0,b_i>0,i=1,2,\dots,k,$ and $ k,n\in N,k\geq1,$ then: $ \frac{{a_1}^{n+1}}{b_1^n}+\frac{a_2^{n+1}}{b_2^n}+\cdots+\frac{{a_k}^{n+1}}{b_k^n}\geq\frac{(a_1+\cdots+a_k)^{n+1}}{(b_1+\cdots+b_k)^n}$ This inequality is right for n any positive real !

$a+b+c=x+y+z\implies \frac{a^{3}}{x^{2}}+x+x+\frac{b^{3}}{y^{2}}+y+y+\frac{c^{3}}{z^{2}}+z+z\geq 3(a+b+c)$ and Then easy by AM-GM $\sum_{cyc}{(\frac{a^{3}}{x^{2}}+x+x)}\geq \sum_{cyc}{\sqrt[3]{\frac{a^3}{x^2}\cdot\ x\cdot\ x}=3a+3b+3c}$

It was also in district olympiad of Kazakhstan in 2010 january for 9 grades. Then this problem was again in city (regional) olympiad in march of 2010 again for 9 grades . It was interesting to see one problem twice in two consecutive rounds.

$\frac{a^3}{x^2}+\frac{b^3}{y^2}+\frac{c^3}{z^2} \ge \frac{(a+b+c)^3}{(x+y+z)^2} =a+b+c$

We have : $\frac{a^3}{x^2}+x+x\geq3a $ (by Cauchy ) Similar : $\frac{b^3}{y^2}+y+y\geq 3b $ and $\frac{c^3}{z^2}+z+z\geq 3c $ => $\frac{a^3}{x^2}+\frac{b^3}{y^2}+\frac{c^3}{z^2}\geq 3(a+b+c)-2(x+y+z)=a+b+c$ ( too easy for a TST problem ) . Expanding problem : $\frac{x^{n+1}}{a^n}+\frac{b^{n+1}}{b^n}+\frac{z^{n+1}}{c^n}\geq a+b+c$ .

By Cauchy Schwarz, $(\frac{a^3}{x^2}+\frac{b^3}{y^2}+\frac{c^3}{z^2})(a+b+c)\geq (\frac{a^2}{x}+\frac{b^2}{y}+\frac{c^2}{z})^2$ and $(\frac{a^2}{x}+\frac{b^2}{y}+\frac{c^2}{z})(x+y+z)\geq (a+b+c)^2$ $\iff$ $\frac{a^2}{x}+\frac{b^2}{y}+\frac{c^2}{z}\geq a+b+c$ Thus we're done

splendid solutions

One step by Titu's Lemma

achen29 wrote: One step by Titu's Lemma Thanks, for your advice about this trivial inequality. Of course, there were no any analogous solutions before. THANK YOU!!!