Let $a,b,c$ be positive real numbers such that $a+b+c=1$. Prove that $$\frac{a}{a^2+b^3+c^3}+\frac{b}{b^2+a^3+c^3}+\frac{c}{c^2+a^3+b^3}\leq\frac{1}{5abc}$$
Problem
Source: RMO 2019 P3
Tags: Inequality, inequalities, am gm
20.10.2019 13:41
Have you done it
20.10.2019 13:53
Homogenise denominator of LHS, AM GM on each denominator after that, multiply entire LHS by the $abc$ in the denominator of RHS, and we get an ineq trivialised by Muirhead on (1,0,0) majorising (3/5,1/5,1/5). One can also finish the last part using weighted AM-GM.
20.10.2019 14:00
I have used Power mean and Jensen
20.10.2019 14:05
I used AM GM everywhere, nothing else
20.10.2019 14:08
How have you homogenized
20.10.2019 14:18
i was not able to do it please some tell me the solution
20.10.2019 14:21
how to do this with AM GM
20.10.2019 14:22
Can someone tell how did they homogenize? @above it's a VERY long bash I suppose
20.10.2019 14:25
Not at all. Just write the 1 in the numerator if RHS$=a+b+c$ and multiply the $a^2$ in denominator of LHS by $1=a+b+c$. It's barely one page using AM GM only, and I have huge handwriting.
20.10.2019 14:33
Math-wiz wrote: I have used Power mean and Jensen Give your solution
20.10.2019 14:51
20.10.2019 15:11
\[(a^2+ab+ac+b^3/a+c^3/a)(3a+b+c) \ge 5 \sqrt[5]{a^2 b^4 c^4} \times 5 \sqrt[5]{a^3 b c}\]\[\implies \sum_{cyc} \frac{a}{a^2+b^3+c^3} \le \sum_{cyc} \frac{3 a + b + c}{25 abc} = \frac{1}{5 abc}\]
20.10.2019 15:19
Oh man! I am soooo dumb! And I had lost hope of homogenising it!
20.10.2019 15:34
20.10.2019 15:43
Arhaan wrote:
How did you arrive at the 3rd line from the 2nd line?
20.10.2019 15:45
My solution, very similar to @5above, @6above but still posting: $$a^2+b^3+c^3=a^2(a+b+c)+b^3+c^3 = a^3+b^3+c^3+a^2b+a^2c \ge 5\sqrt[5]{a^7b^4c^4}$$$$\implies \sum_{cyc} \frac{a^2bc}{a^2+b^3+c^3} \le \sum_{cyc} \frac{\sqrt[5]{a^3bc}}{5} \le \frac{a+b+c}{5} = \frac15$$$$\implies \sum_{cyc} \frac{a}{a^2+b^3+c^3} \le \frac{1}{5abc}$$The last step can be done using Muirhead on $(1,0,0)$ and $(\frac35,\frac15,\frac15)$, or since this is RMO: $$a+a+a+b+c \ge 5\sqrt[5]{a^3bc}$$Now sum cyclically and finish. Oh drat identical to @2above
20.10.2019 15:59
RamK wrote:
I'm sorry that was a typo. There was supposed to be reciprocal of all the terms in the denominator.
20.10.2019 16:48
Let $a,b,c$ be positive real numbers such that $a+b+c=1$. Prove that $$\frac{a}{a+b^2+c^2}+\frac{b}{b+a^2+c^2}+\frac{c}{c+a^2+b^2}\leq\frac{1}{5}\left(\frac{a}{bc}+\frac{b}{ca}+\frac{c}{ab}\right).$$here
20.10.2019 17:17
I am feeling the same as Tom Holland did ok that bridge after that hologram thing!
30.10.2019 23:01
Any guesses of the cut-off?
16.05.2020 11:44
Just curious, can this be done by Denominator Nullification?
30.05.2020 19:56
By Jensen , I did it in 5 steps
07.06.2020 16:25
Kgxtixigct wrote: By Jensen , I did it in 5 steps How did you do that please post...
08.10.2020 22:44
After a year, we made truces for no reason $\sum_{cyc} \dfrac{a}{a^2(a+b+c)+b^3+c^3} \leq \sum_{cyc} \dfrac{1}{a^2 + ab + ab + \dfrac{b^3}{a}+\dfrac{c^3}{a}} \overset{GM-HM}{\leq} $ $\sum_{cyc} \dfrac{(a^3bc)^5}{5abc} \overset{wAM-GM}{\leq} \sum_{cyc} \dfrac{3a+b+c}{5abc} = \dfrac{1}{5abc}$. The n00bier days.
03.11.2020 10:30
Can anyone tell which function to consider to finish this problem using Jensen's inequality??? Is there any solution purely by Jensen?
29.08.2021 05:08
$$\frac{a}{a^2+b^3+c^3}+\frac{b}{b^2+a^3+c^3}+\frac{c}{c^2+ a^3+b^3}\leq\frac{1}{5abc}$$
22.12.2021 17:16
Let $a,b,c$ be positive real numbers such that $a+b+c=1$. Prove that $$\frac{a}{a^2+b+c}+\frac{b}{b^2+c+a}+\frac{c}{c^2+a+b}\leq\frac{1}{21abc}$$
23.12.2021 02:41
Let $a,b,c$ be positive real numbers such that $a+b+c=1$. Prove that $$\frac{a}{a^3+b+c}+\frac{b}{b^3+c+a}+\frac{c}{c^3+a+b}\leq\frac{1}{19abc}$$
23.12.2021 11:40
sqing wrote: Let $a,b,c$ be positive real numbers such that $a+b+c=1$. Prove that $$\frac{a}{a+b^2+c^2}+\frac{b}{b+a^2+c^2}+\frac{c}{c+a^2+b^2}\leq\frac{1}{15abc}.$$
Attachments:

23.12.2021 15:44
Let $a,b,c$ be positive real numbers such that $a+b+c=1$. Prove that $$\frac{a^2}{a+b^2+c^2}+\frac{b^2}{b+a^2+c^2}+\frac{c^2}{c+a^2+b^2}\leq\frac{1}{45abc}$$$$\frac{a^2}{a^2+b+c}+\frac{b^2}{b^2+c+a}+\frac{c^2}{c^2+a+b}\leq\frac{1}{63abc}$$$$ \frac{a^2}{a^2+b^3+c^3}+\frac{b^2}{b^2+a^3+c^3}+\frac{c^2}{c^2+ a^3+b^3}\leq\frac{1}{15abc}$$$$ \frac{a^3}{a^2+b^3+c^3}+\frac{b^3}{b^2+a^3+c^3}+\frac{c^3}{c^2+ a^3+b^3}\leq\frac{1}{45abc}$$
09.10.2023 23:31
We can write $\frac{a}{a^2+b^3+c^3}\\ = \frac{a}{a^2+(b+c)(b^2+c^2-bc)}\\ = \frac{a}{a^2+(1-a)((1-a)^2-3\frac{abc}{a})}$ We can now notice that if $abc$ were to be maximized, then the LHS will be at it's maxima and LHS at it's minima. If the inequality holds true then, it does hold for all other conditions as we are not fixing $a,b,c$ only substituting $abc=\frac{1}{27}$. The inequality now is: $\sum f(a)= \sum \frac{a}{a^2+(1-a)^3-\frac{1-a}{9a})} \leq \frac{27}{5}$ We will now use Jensen to claim that $\sum f(a)\leq 3f(\frac{a+b+c}{3}) = 3f(\frac{1}{3})=\frac{27}{5}$. Hence, proved.
21.10.2023 08:13
11.09.2024 16:43
Solution: Homogenise to get $$\sum_cyc \frac{a}{a^3 + b^3 + c^3 + a^2b + a^2c} \le \frac{a+b+c}{5abc}.$$Now we have that $$\frac{a^3 + b^3 + c^3 + a^2b + a^2c}{5} \ge a^{\frac 75} b^{ \frac 45} c^{\frac 45}$$from AM-GM, so $$ \frac{a}{a^3 + b^3 + c^3 + a^2b + a^2c} \le \frac{a^{\frac 35} b^{\frac 15} c^{\frac 15}}{5abc}$$. Therefore it suffices to show that $$\sum_{cyc} a^{\frac 35} b^{\frac 15} c^{\frac 15} \le a+b+c \iff \sum_{sym} a^{\frac 35} b^{\frac 15} c^{\frac 15} \le \sum_{sym} a^1b^0c^0,$$true by Muirhead.
26.11.2024 13:15
Loved this ineq.