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.

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?

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. -_-

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?

Probably the intended solution:

My solution:

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.

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...

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.

I just used Titu's Lemma and used that $a^2+b^2+c^2 \ge ab+bc+ca$.

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.

yea i multiplied it out and used muirhead's 6 times, but hey that's why you have 4 1/2 hours

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.

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: \[{\sum \frac{a^4}{5a^2 + ab} + 3 \sum \frac{b^4}{5ab + b^2} \ge \frac{9}{15 + ab+bc+ca} + \frac{27}{5(ab+bc+ca)+3} \ge 2 \iff ab+bc+ca \le 3._{\blacksquare}}\]

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

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*}

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.

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)

multiply by a/a and by b/b

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.

its so easy

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.

$\frac{a^3+3b^3}{5a+b}+\frac{b^3+3c^3}{5b+c}+\frac{c^3+3a^3}{5c+a} \ge \frac{2a^2+2b^2+2c^2}{3}$ $(\frac{a^3}{5a+b}+\frac{b^3}{5b+c}+\frac{c^3}{5c+a})+(\frac{3b^3}{5a+b}+\frac{3c^3}{5b+c}+\frac{3a^3}{5c+a}) \ge \frac{2a^2+2b^2+2c^2}{3}$ $(\frac{a^4}{5a^2+ab}+\frac{b^4}{5b^2+bc}+\frac{c^4}{5c^2+ac})+(\frac{3b^4}{5ab+b^2}+\frac{3c^4}{5bc+c^2}+\frac{3a^4}{5ac+a^2}) \ge \frac{2a^2+2b^2+2c^2}{3}$ 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 $(\frac{(a^2+b^2+c^2)^2}{5a^2+5b^2+5c^2+ab+bc+ac})+(\frac{3(a^2+b^2+c^2)^2}{a^2+b^2+c^2+5ab+5bc+5ac}) \ge (\frac{(a^2+b^2+c^2)^2}{6a^2+6b^2+6c^2})+(\frac{3(a^2+b^2+c^2)^2}{6a^2+6b^2+6c^2})=\frac{2a^2+2b^2+2c^2}{3}$

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.

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. $\square$

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.

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 $\square$

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 $a^2+b^2+c^2 \geq ab+ac+bc$. $\blacksquare$

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.

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 $$\sum_{cyc}\frac{a^3+3b^3}{5a+b} = \sum_{cyc}\frac{a^3}{5a+b} + \sum_{cyc}\frac{3b^3}{5a+b}$$$$\geq (a^2+b^2+c^2)(\frac{1}{2}+\frac{1}{6} = \frac{2}{3}(a^2+b^2+c^2)$$

I discussed this problem in my youtube channel (little fermat) Video in my inequalities tutorial playlist.

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.

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 $$\sum_{cyc} (\frac{a^4}{5a^2 + ab} + \frac{3b^4}{5ab + b^2}) \geq \frac{2}{3} (a^2 + b^2 + c^2). $$

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, $$\frac{a^4}{5a^2+ab}+\frac{b^4}{5b^2+bc}+\frac{c^4}{5c^2+ac}+3\cdot(\frac{a^4}{5ca+a^2}+\frac{b^4}{5ab+b^2}+\frac{c^4}{5bc+c^2})$$$$\ge \frac{(a^2+b^2+c^2)^2}{5a^2+5b^2+5c^2+ab+bc+ca}+3\cdot\frac{(a^2+b^2+c^2)^2}{5ab+5bc+5ca+a^2+b^2+c^2}$$$$\ge \frac{(a^2+b^2+c^2)^2}{6a^2+6b^2+6c^2}+3\cdot\frac{(a^2+b^2+c^2)^2}{6a^2+6b^2+6c^2}$$$$=\frac{2}{3}(a^2+b^2+c^2).$$