Let $a,b,c$ be positive real numbers. Prove that $\frac{a^3+3b^3}{5a+b}+\frac{b^3+3c^3}{5b+c}+\frac{c^3+3a^3}{5c+a} \geq \frac{2}{3}(a^2+b^2+c^2)$.
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Tags: AMC, USA(J)MO, USAJMO, inequalities, algebra, polynomial, ratio
25.04.2012 01:05
0. >.< My friend said you needed to prove $\frac{a^3+3b^3}{5a+b} \ge \frac{1}{2}a^2+\frac{1}{6}b^2$ or something like that.
25.04.2012 01:06
I showed that when $a = b = c$ we have LHS = RHS Then I set $a+b+c = x$ and showed that as $c$ approaches $x$ and $a$ and $b$ approach $0$, we have that $\frac{16}{5} \geq \frac{2}{3}$ Only, I'm not sure whether I wrote $c$ approaches $x$ or infinity... Then I got clueless, so I guessed "It is easy to show that for any $a,b,c$ between these two extremes, LHS is still greater than or equal to RHS, QED. AkshajK wrote: How many points do you think i'll get?
25.04.2012 01:08
I got the inequality to $\sum_{cyc} \frac{a^3+3b^3}{5a+b}\ge\frac{6abc}{a+b+c}$ using AM-GM, but just bsed the rest to show the desired. -_-
25.04.2012 01:10
Neither of those looks like it'll get a positive score :3 So I wrote "By rearrangement inequality on {a,b,c} and {a^2,b^2,c^2}, we have $a^3+b^3+c^3 \geq ab^2+bc^2+ca^2$. Thus, $12a^3+12b^3+12c^2 \geq 10a^3+10b^3+10c^2+2(ab^2+bc^2+ca^2)$ Thus, $(a^3+3b^3)+(b^3+3c^3)+(c^3+3a^3) \geq \frac{2}{3}a^2(5a+b)+\frac{2}{3}b^2(5b+c)+\frac{2}{3}c^2(5c+a)$" And stopped. Obviously this doesn't do a whole lot (well, nothing). Any possibility of sneaking a point?
25.04.2012 01:10
Probably the intended solution:
My solution:
25.04.2012 01:13
Well, what I did is that I didn't really know what to do after an hour of bashing and stuff, so I used Cauchy, used a faulty muirhead (flipped the inequality sign ), and proved that. Hopefully they won't notice... or read this post.
25.04.2012 01:30
By the way, I didn't take the JMO, but this took me about 20 minutes. I guess I probably should have tried Cauchy first, as above...
25.04.2012 01:32
Is there anyway to legitly solve this with Nesbitt's? because i saw the $\frac{2}{3}$, flipped it and that was the motivation for my bs.
25.04.2012 02:23
I just used Titu's Lemma and used that $a^2+b^2+c^2 \ge ab+bc+ca$.
25.04.2012 02:53
How much credit would they give if I said there exists an x such that (a^3+3b^3)/(5a+b)>=x*a^2+(2/3-x)b^2, and this is shown when letting q=(a/b), we have x*f(q)>=g(q) for all positives q, where f, g are quartic polynomials, and since it's well known that there's always a global maximum for the ratio of two polynomials with the same degree, x exists? Basically, this seems like the motivation for a well known technique (which I didn't know), where you use weighted am-gm a bunch.
25.04.2012 03:01
yea i multiplied it out and used muirhead's 6 times, but hey that's why you have 4 1/2 hours
25.04.2012 03:06
It shouldn't take more than 1 hour with Muirhead if you know what you're doing. This problem screams of Muirhead, considering that both sides have the same degree, and multiplying will get 1296 terms on each side.
25.04.2012 03:13
lol solved this in a minute literally.. but I didn't take the test : just WLOG $a^2 + b^2 + c^2 = 3$ and use CS: \[{\sum \frac{a^4}{5a^2 + ab} + 3 \sum \frac{b^4}{5ab + b^2} \ge \frac{9}{15 + ab+bc+ca} + \frac{27}{5(ab+bc+ca)+3} \ge 2 \iff ab+bc+ca \le 3._{\blacksquare}}\]
25.04.2012 03:13
2 muirheads could have been enough but more doesn't hurt, i still failed to solve problem 1 in the 2 1/2 hours i spent on it cause im terrible at geo
25.04.2012 03:47
From Cauchy-Schwarz (Titu's Lemma), \begin{align*}\sum_{cyc} \frac{a^3}{5a+b} + \sum_{cyc} \frac{3a^3}{5c+a} &=\sum_{cyc} \frac{a^4}{5a^2+ab} + 3 \sum_{cyc} \frac{a^4}{5ac+a^2} \\&\ge \frac{(a^2+b^2+c^2)^2}{5(a^2+b^2+c^2)+ab+bc+ca} + 3\cdot \frac{(a^2+b^2+c^2)^2}{a^2+b^2+c^2+5(ab+bc+ca)} \\&\ge \frac{a^2+b^2+c^2}{6} + \frac{a^2+b^2+c^2}{2} \\&=\frac{2}{3}(a^2+b^2+c^2)\end{align*}
25.04.2012 03:59
I didn't take the Jmo but this problem took me exactly 15 seconds to do so yeah... I think it's due to hanging around the inequality marathon too long.
25.04.2012 05:49
proglote wrote: lol solved this in a minute literally.. but I didn't take the test : just WLOG $a^2 + b^2 + c^2 = 3$ and use CS: \[{\sum \frac{a^4}{5a^2 + ab} + 3 \sum \frac{b^4}{5ab + b^2} \ge \frac{9}{15 + ab+bc+ca} + \frac{27}{5(ab+bc+ca)+3} \ge 2 \iff ab+bc+ca \le 3._{\blacksquare}}\] Why is it a fourth power? a^4?, is this cauchy after all simplification? Can you go step through step please, I got far on my own with this problem but I dont see it as fast as some did.(didnt write jmo or usamo)
25.04.2012 07:02
multiply by a/a and by b/b
25.04.2012 13:31
I do this by showing $\frac{a^3+b^3}{5a+b} \geq \frac{1}{36}(25b^2-a^2)$ which is equivalent to $(a-b)^2(41a+83b)\geq 0$. The rest follows quite easily from here.
09.08.2023 15:11
its so easy
28.09.2023 06:25
Consider Titu's which gives, \[ \sum_{cyc} \frac{a^3 + 3b^3}{5a + b} \geq \sum_{cyc} \frac{a^4}{5a^2+ab} + \frac{3b^4}{5ab + b^2} \geq \frac{(a^2+b^2+c^2)^2}{\sum_{cyc} 5a^2 + \sum_{cyc} ab} + \frac{3(a^2+b^2+c^2)^2}{\sum_{cyc} a^2 + \sum_{cyc} 5ab} \]Now we wish to show, \begin{align*} \frac{(a^2+b^2+c^2)^2}{\sum_{cyc} 5a^2 + \sum_{cyc} ab} + \frac{3(a^2+b^2+c^2)^2}{\sum_{cyc} a^2 + \sum_{cyc} 5ab} &\geq \frac{2}{3} \cdot (a^2 + b^2 + c^2)\\ \\ \frac{(a^2+b^2+c^2)}{\sum_{cyc} 5a^2 + \sum_{cyc} ab} + \frac{3(a^2+b^2+c^2)}{\sum_{cyc} a^2 + \sum_{cyc} 5ab} &\geq \frac{2}{3}\\ \\ \frac{(a^2+b^2+c^2)}{\sum_{cyc} 5a^2 + \sum_{cyc} ab} + \frac{9(a^2+b^2+c^2)}{\sum_{cyc} 3a^2 + \sum_{cyc} 15ab} &\geq \frac{2}{3}. \end{align*}Now Titu's again gives, \[\frac{1}{\sum_{cyc} 5a^2 + \sum_{cyc} ab} + \frac{9}{\sum_{cyc} 3a^2 + \sum_{cyc} 15ab} \geq \frac{16}{8(a+b+c)^2}\]Then we can simplify the equation to showing, \begin{align*} 3(a^2+b^2+c^2) &\geq (a+b+c)^2\\ a^2+b^2+c^2 &\geq ab +bc+ca, \end{align*}so we are done by Rearrangement.
10.10.2023 18:32
$\frac{a^3+3b^3}{5a+b}+\frac{b^3+3c^3}{5b+c}+\frac{c^3+3a^3}{5c+a} \ge \frac{2a^2+2b^2+2c^2}{3}$ $(\frac{a^3}{5a+b}+\frac{b^3}{5b+c}+\frac{c^3}{5c+a})+(\frac{3b^3}{5a+b}+\frac{3c^3}{5b+c}+\frac{3a^3}{5c+a}) \ge \frac{2a^2+2b^2+2c^2}{3}$ $(\frac{a^4}{5a^2+ab}+\frac{b^4}{5b^2+bc}+\frac{c^4}{5c^2+ac})+(\frac{3b^4}{5ab+b^2}+\frac{3c^4}{5bc+c^2}+\frac{3a^4}{5ac+a^2}) \ge \frac{2a^2+2b^2+2c^2}{3}$ By Titu's lemma, $(\frac{a^4}{5a^2+ab}+\frac{b^4}{5b^2+bc}+\frac{c^4}{5c^2+ac})+(\frac{3b^4}{5ab+b^2}+\frac{3c^4}{5bc+c^2}+\frac{3a^4}{5ac+a^2}) \ge (\frac{(a^2+b^2+c^2)^2}{5a^2+5b^2+5c^2+ab+bc+ac})+(\frac{3(a^2+b^2+c^2)^2}{a^2+b^2+c^2+5ab+5bc+5ac})$ $a^2+b^2+c^2 \ge ab+bc+ac$ because $(2,0,0) > (1,1,0)$ by muirhead So $(\frac{(a^2+b^2+c^2)^2}{5a^2+5b^2+5c^2+ab+bc+ac})+(\frac{3(a^2+b^2+c^2)^2}{a^2+b^2+c^2+5ab+5bc+5ac}) \ge (\frac{(a^2+b^2+c^2)^2}{6a^2+6b^2+6c^2})+(\frac{3(a^2+b^2+c^2)^2}{6a^2+6b^2+6c^2})=\frac{2a^2+2b^2+2c^2}{3}$
15.10.2023 03:43
It suffices to show: \[\sum{\text{cyc}}\frac{a^3}{5a+b}+3\sum{\text{cyc}}\frac{b^3}{5a+b}\geq\frac{2}{3}(a^2+b^2+c^2)\] By Titu's Lemma, the LHS is greater than or equal to: \[\frac{(a^2+b^2+c^2)^2}{\sum_{\text{cyc}}5a^2+\sum_{\text{cyc}}ab}+\frac{3(a^2+b^2+c^2)^2}{\sum_{\text{cyc}}5ab+\sum_{\text{cyc}}a^2}.\]Thus, it suffices to show: \[\frac{a^2+b^2+c^2}{\sum_{\text{cyc}}5a^2+\sum_{\text{cyc}}ab}+\frac{3(a^2+b^2+c^2)}{\sum_{\text{cyc}}5ab+\sum_{\text{cyc}}a^2}\geq \frac{2}{3}.\] Applying Titu's Lemma again, this is greater than or equal to: \[\frac{(4a)^2+(4b)^2+(4c)^2}{\sum_{\text{cyc}}8a^2+\sum_{\text{cyc}}16ab}=\frac{2a^2+2b^2+2c^2}{\sum_{\text{cyc}}a^2+\sum_{\text{cyc}}2ab}.\] Expanding, it suffices to show: \[3\sum_{\text{sym}}a^2\geq\sum_{\text{sym}}a^2+2\sum_{\text{sym}}ab,\]\[2\sum_{\text{sym}}a^2\geq 2\sum_{\text{sym}}ab.\]As $(2,0,0)\succ (1,1,0)$, the above inequality is true by Murihead's inequality.
24.10.2023 20:50
Notice \begin{align*} \sum_{\text{cyc}} \frac{a^3}{5a+b} &= \sum_{\text{cyc}} \frac{a^4}{5a^2+ab} \\ &\ge \frac{(a^2+b^2+c^2)^2}{5a^2+5b^2+5c^2+ab+bc+ca} \\ & \ge \frac{a^2+b^2+c^2}{6} \end{align*} Similarly, we obtain \begin{align*} \sum_{\text{cyc}} \frac{3b^3}{5a+b} &= \sum_{\text{cyc}} \frac{3b^4}{5ab+b^2} \\ & \ge \frac{3(a^2+b^2+c^2)^2}{5ab+5bc+5ca+a^2+b^2+c^2} \\ & \ge \frac{a^2+b^2+c^2}{2} \end{align*} Adding gives the desired results. $\square$
26.11.2023 08:16
By Titu, \begin{align*} \sum_{\text{cyc}} \left(\frac{a^4}{a(5a+b)}+\frac{3b^4}{b(5a+b)}\right) &\ge \frac{(a^2+b^2+c^2)^2}{\sum_{\text{cyc}} (5a^2+ab)}+\frac{3(a^2+b^2+c^2)^2}{\sum_{\text{cyc}} (5ab+b^2)}, \end{align*}so it suffices to show that \begin{align*} \frac{a^2+b^2+c^2}{\sum_{\text{cyc}} (5a^2+ab)}+\frac{3(a^2+b^2+c^2)}{\sum_{\text{cyc}} (5ab+b^2)} &\ge \frac23. \end{align*}However, since $a^2+b^2+c^2 \ge ab+bc+ca$, the inequality becomes \begin{align*} \frac{a^2+b^2+c^2}{6(a^2+b^2+c^2)}+\frac{3(a^2+b^2+c^2)}{6(a^2+b^2+c^2)} &\ge \frac23, \end{align*}which is clear after simplification.
24.01.2024 05:27
Inequalities are fun . By Titu's Lemma $$\left( \sum_{\text{cyc}} \frac{a^{3}}{5a^{2}+ab}+3 \cdot \frac{b^{4}}{5ab+b^{2}} \right) \geq \frac{16(a^{2}+b^{2}+c^{2})^{2}}{8(a^{2}+b^{2}+c^{2})+16(ab+ac+bc)}= \frac{2(a^{2}+b^{2}+c^{2})^{2}}{(a+b+c)^{2}}$$Now it would suffice to show $$\frac{2(a^{2}+b^{2}+c^{2})^{2}}{(a+b+c)^{2}} \geq \frac{2}{3}(a^2+b^2+c^2)$$Or $$3a^{2}+3b^{2}+3c^{2} \geq (a+b+c)^{2}$$Expansion and rearrangement gives it suffices to show: $$(a-b)^{2}+(b-c)^{2}+(c-a)^{2} \geq 0$$Which is true by the trivial inequality $\square$
26.02.2024 04:48
We let $\frac{a^3+3b^3}{5a+b} = \frac{a^4}{a(5a+b)} + \frac{3b^4}{b(5a+b)}$. Applying Titu's lemma we get \[\sum_{\text{cyc}} \frac{a^4}{a(5a+b)} + \frac{3b^4}{b(5a+b)} \geq \frac{(a^2+b^2+c^2)^2}{5a^2+5b^2+5c^2+ab+ac+bc} + \frac{(a^2+b^2+c^2)^2}{a^2+b^2+c^2+5ab+5ac+5bc}\]Thus we just need to prove that \[\frac{(a^2+b^2+c^2)}{5a^2+5b^2+5c^2+ab+ac+bc} + \frac{(a^2+b^2+c^2)}{a^2+b^2+c^2+5ab+5ac+5bc} \geq \frac{2}{3}\]This is obvious since $a^2+b^2+c^2 \geq ab+ac+bc$. $\blacksquare$
06.09.2024 06:27
Split the fractions apart and apply Titu's on the two triples to reduce the inequality into: $$\frac{(a^2+b^2+c^2)^2}{5(a^2+b^2+c^2)+(ab+bc+ac)} + \frac{3(a^2+b^2+c^2)^2}{a^2+b^2+c^2+5(ab+bc+ac)} \geq RHS$$or by cancelling the ugly variable term on the RHS: $$\frac{a^2+b^2+c^2}{5(a^2+b^2+c^2)+ab+bc+ac} + \frac{3(a^2+b^2+c^2)}{a^2+b^2+c^2+5(ab+bc+ac)} \geq \frac23$$Sub $a^2+b^2+c^2=x$ and $ab+bc+ac=y$ for brevity and simplift further: $$\frac{x}{5x+y} + \frac{3x}{x+5y} \geq \frac23$$$$38x^2 \geq 28xy+10y^2$$$$38 \left( \frac{x}{y}\right)^2 \geq 28 \left( \frac{x}{y} \right) + 10$$so we have reduced the inequality into $x \geq y$ which is trivial.
14.09.2024 11:43
We prove the inequality by spitting sum into two easier sums and then finishing by using holder's inequalty and the following am-gm application $a^2+b^2+c^2 \geq ab+bc+ca$ $$(\sum_{cyc}\frac{a^3}{5a+b})(\sum_{cyc}a(5a+b)) \geq (a^2+b^2+c^2)^2$$$$\sum_{cyc}\frac{a^3}{5a+b} \geq \frac{(a^2+b^2+c^2)^2}{5a^2+5b^2+5c^2+ab+bc+ca} $$$$ \geq\frac{(a^2+b^2+c^2)^2}{6(a^2+b^2+c^2)} = \frac{a^2+b^2+c^2}{6} $$and similarly using holder's inequality and am-gm for the remaining sum we get $$\sum_{cyc}\frac{3b^3}{5a+b}\geq \frac{3(a^2+b^2+c^2)^2}{a^2+b^2+c^2+5ab+5bc+5ca}$$$$\geq \frac{3(a^2+b^2+c^2)^2}{6(a^2+b^2+c^2)} = \frac{a^2+b^2+c^2}{2}$$and finallyusing the above two inequalities we get $$\sum_{cyc}\frac{a^3+3b^3}{5a+b} = \sum_{cyc}\frac{a^3}{5a+b} + \sum_{cyc}\frac{3b^3}{5a+b}$$$$\geq (a^2+b^2+c^2)(\frac{1}{2}+\frac{1}{6} = \frac{2}{3}(a^2+b^2+c^2)$$
16.09.2024 22:32
I discussed this problem in my youtube channel (little fermat) Video in my inequalities tutorial playlist.
19.09.2024 18:43
Simply note that from Titu's Lemma we have, \[\frac{a^3}{5a+b}+\frac{b^3}{5b+c}+\frac{c^3}{5c+a} = \frac{a^4}{5a^2+ab}+\frac{b^4}{5b^2+bc}+\frac{c^4}{5c^2+ca} \ge \frac{(a^2+b^2+c^2)^2}{5(a^2+b^2+c^2)+ab+c+ca}\]However, by the Rearrangement inequality we know, $a^2+b^2+c^2\ge ab+bc+ca$. Thus, \[\frac{a^3}{5a+b}+\frac{b^3}{5b+c}+\frac{c^3}{5c+a} \ge \frac{(a^2+b^2+c^2)^2}{5(a^2+b^2+c^2)+ab+c+ca} \ge \frac{(a^2+b^2+c^2)^2}{6(a^2+b^2+c^2)}=\frac{1}{6}(a^2+b^2+c^2)\]A similar argument also shows, \[\frac{3b^3}{5a+b}+\frac{3c^3}{5b+c}+\frac{3a^3}{5c+a} \ge \frac{1}{2}(a^2+b^2+c^2)\]Summing these inequality now yields, \[\frac{a^3+3b^3}{5a+b}+\frac{b^3+3c^3}{5b+c}+\frac{c^3+3a^3}{5c+a} \ge \frac{1}{6}(a^2+b^2+c^2)+\frac{1}{2}(a^2+b^2+c^2)=\frac{2}{3}(a^2+b^2+c^2)\]which finishes the proof.
22.09.2024 07:15
Note that: As $(2,0,0) \succ (1,1,0)$ by Muirhead's Inequality, , $$ a^2 + b^2 + c^2 \geq ab + bc +ca$$We can split the R.H.S. of the original equation as follows $$\sum_{cyc} (\frac{a^4}{5a^2 + ab} + \frac{3b^4}{5ab + b^2})$$Applying Titu's Lemma separately , $$\begin{aligned} \sum_{cyc} \frac{a^4}{5a^2 + ab} &\geq \frac{(a^4+b^4+c^4)^2}{5( a^2 + b^2 + c^2) + ab + bc + ca}\\ &\geq \frac{(a^2+b^2+c^2)^2}{6( a^2 + b^2 + c^2)}\\ &\geq \frac{(a^2+b^2+c^2)}{6} \end{aligned}$$$$\begin{aligned} \sum_{cyc} \frac{3b^4}{5ab + b^2} &\geq \frac{3(a^4+b^4+c^4)^2}{a^2 + b^2 + c^2 + 5(ab + bc + ca)}\\ &\geq \frac{3(a^2+b^2+c^2)^2}{6( a^2 + b^2 + c^2)}\\ &\geq \frac{(a^2+b^2+c^2)}{2} \end{aligned}$$Summing up the 2 inequalities $$\sum_{cyc} (\frac{a^4}{5a^2 + ab} + \frac{3b^4}{5ab + b^2}) \geq \frac{2}{3} (a^2 + b^2 + c^2). $$
03.11.2024 17:15
We want to prove that $$\frac{a^3}{5a+b}+\frac{b^3}{5b+c}+\frac{c^3}{5c+a}+3\cdot(\frac{a^3}{5c+a}+\frac{b^3}{5a+b}+\frac{c^3}{5b+c})\ge \frac{2}{3}(a^2+b^2+c^2).$$By Titu's Lemma, $$\frac{a^4}{5a^2+ab}+\frac{b^4}{5b^2+bc}+\frac{c^4}{5c^2+ac}+3\cdot(\frac{a^4}{5ca+a^2}+\frac{b^4}{5ab+b^2}+\frac{c^4}{5bc+c^2})$$$$\ge \frac{(a^2+b^2+c^2)^2}{5a^2+5b^2+5c^2+ab+bc+ca}+3\cdot\frac{(a^2+b^2+c^2)^2}{5ab+5bc+5ca+a^2+b^2+c^2}$$$$\ge \frac{(a^2+b^2+c^2)^2}{6a^2+6b^2+6c^2}+3\cdot\frac{(a^2+b^2+c^2)^2}{6a^2+6b^2+6c^2}$$$$=\frac{2}{3}(a^2+b^2+c^2).$$