Let $a$, $b$, $c$ be positive reals. Prove that \[ \sqrt{\frac{a^2(bc+a^2)}{b^2+c^2}}+\sqrt{\frac{b^2(ca+b^2)}{c^2+a^2}}+\sqrt{\frac{c^2(ab+c^2)}{a^2+b^2}}\ge a+b+c. \]Proposed by Robin Park
Problem
Source: ELMO 2014 Shortlist A9, by Robin Park
Tags: inequalities, inequalities proposed
arqady
24.07.2014 21:20
v_Enhance wrote: Let $a$, $b$, $c$ be positive reals. Prove that \[ \sqrt{\frac{a^2(bc+a^2)}{b^2+c^2}}+\sqrt{\frac{b^2(ca+b^2)}{c^2+a^2}}+\sqrt{\frac{c^2(ab+c^2)}{a^2+b^2}}\ge a+b+c. \]Proposed by Robin Park Let $a\geq b\geq c$. Hence, $\sum_{cyc}(\sqrt{\frac{a^2(bc+a^2)}{b^2+c^2}}-a)=\sum_{cyc}(\frac{2a(bc+a^2)}{2\sqrt{(bc+a^2)(b^2+c^2)}}-a)\geq$ ${\geq\sum_{cyc}(\frac{2a(bc+a^2)}{a^2+b^2+c^2+bc}}-a)=\sum_{cyc}\frac{(a-b)(a^2+2ab-ac)-(c-a)(a^2+2ac-ab)}{2(a^2+b^2+c^2+bc)}=$ $=\sum_{cyc}(a-b)(\frac{a^2+2ab-ac}{2(a^2+b^2+c^2+bc)}-\frac{b^2+2ab-bc}{2(a^2+b^2+c^2+ac)})=$ $=\sum_{cyc}\frac{(a-b)^2(a^3+a^2b+ab^2+b^3+2abc-c^3)(a^2+b^2+c^2+ab)}{2\prod\limits_{cyc}(a^2+b^2+c^2+ab)}\geq$ $\geq\sum_{cyc}\frac{(a-b)^2(a^3+b^3-c^3)(a^2+b^2+c^2+ab)}{2\prod\limits_{cyc}(a^2+b^2+c^2+ab)}\geq$ $\geq\frac{(a-c)^2(a^3+c^3-b^3)(a^2+b^2+c^2+ac)}{2\prod\limits_{cyc}(a^2+b^2+c^2+ab)}+$ $+\sum_{cyc}\frac{(b-c)^2(b^3+c^3-a^3)(a^2+b^2+c^2+bc)}{2\prod\limits_{cyc}(a^2+b^2+c^2+ab)}\geq$ $\geq\frac{(b-c)^2(a^3-b^3)(a^2+b^2+c^2+ac)}{2\prod\limits_{cyc}(a^2+b^2+c^2+ab)}+\frac{(b-c)^2(b^3-a^3)(a^2+b^2+c^2+bc)}{2\prod\limits_{cyc}(a^2+b^2+c^2+ab)}=$ $=\frac{(b-c)^2(a-b)^2(a^2+ab+b^2)c}{2\prod\limits_{cyc}(a^2+b^2+c^2+ab)}\geq0$.
Pain rinnegan
24.07.2014 22:23
It's arqady's inequality! See here.
arqady
25.07.2014 06:22
Thank you, Pain rinnegan! I forgot about it.
BSJL
26.07.2014 10:21
Wow, how a nice proof!! My solution is that~ $ (LHS)^2(\sum{a(a^2+bc)^2(b^2+c^2)}) \ge (\sum{a(a^2+bc)})^3 $ Thus, we suffice to show : $ (\sum{a(a^2+bc)})^3 \ge (a+b+c)^2(\sum{a(a^2+bc)^2(b^2+c^2)}) $ which is true after full expand...
infiniteturtle
05.03.2015 18:09
Another proof with expansion... By Holder, \[(\text{LHS})^2\left(\sum a(b^2+c^2)\right)\left(\sum \frac{a}{a^2+bc}\right)\ge (a+b+c)^4.\]It suffices to show that \[\left(\sum_{\text{sym}} a^2b\right)\left(\sum a(b^2+ca)(c^2+ab)\right)\le (a+b+c)^2(a^2+bc)(b^2+ca)(c^2+ab).\]
We can actually do this pretty cleanly. The idea is to retain use of symmetric sums.
First off, we can expand the $a(b^2+ca)(c^2+ab)$ on the LHS easily. The $(a^2+bc)(b^2+ca)(c^2+ab)$ on the RHS is also easy to deal with; see what we get when we multiply all squares, 2,1, or 0 squares. So we want to show that \[\left(\sum_{\text{sym}} a^2b\right)\left(ab^2c^2+a^2bc^2+a^2b^2c+a^2b^3+a^2c^3+b^2c^3+b^2a^3+c^2a^3+c^2b^3+a^3bc+ab^3c+abc^3\right)\le \]\[\le (a^2+b^2+c^2+2ab+2bc+2ca)\left(\sum_{\text{cyc}}a^3b^3 +\sum_{\text{cyc}}a^4bc+2a^2b^2c^2\right).\]It's faster to use symmetric sums on the RHS, but I like to avoid fractions. Now on the LHS, we can just multiply $a^2b$ by everything and take the symmetric sum of that; on the RHS, we can multiply our expanded $(a+b+c)^2$ by each of the three terms, then group into symmetric sums. After we do this and clean things up, we get \[\sum_{\text{sym}}a^5b^3+\sum_{\text{sym}}2a^5b^2c+\sum_{\text{sym}}2a^4b^2c^2+\sum_{\text{sym}}3a^4b^3c+\sum_{\text{sym}}a^4b^4+\sum_{\text{sym}}3a^3b^3c^2\le\]\[\le \tfrac{1}{2}\sum_{\text{sym}}a^6bc+\sum_{\text{sym}}a^5b^3+\sum_{\text{sym}}2a^5b^2c+\sum_{\text{sym}}2a^4b^2c^2+\sum_{\text{sym}}3a^4b^3c+\sum_{\text{sym}}a^4b^4+\tfrac{5}{2}\sum_{\text{sym}}a^3b^3c^3.\]Miraculously, nearly everything cancels! Leaving \[\tfrac{1}{2}\sum_{\text{sym}}a^6bc\ge \tfrac{1}{2}\sum_{\text{sym}}a^3b^3c^2,\]which is quite clear by Muirhead!
By doing the AM-GM in our Muirhead, it's clear that equality can only be achieved when $a=b=c$ or when $a=0$ (and permutations). Thankfully we don't have to try to analyze equality in the Holder; we can just plug back into the original and see that $a=b=c$ works and $a=0$ reduces to $b^3+c^3\ge b^2c+bc^2$, which has equality at $b=c$. So the equality cases are $(a,a,a),(0,a,a)$ and permutations of the second.
Out of a 144 term mess we end up with 6. We win!