Have you done it

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.

I have used Power mean and Jensen

I used AM GM everywhere, nothing else

How have you homogenized

i was not able to do it please some tell me the solution

how to do this with AM GM

Can someone tell how did they homogenize? @above it's a VERY long bash I suppose

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.

Math-wiz wrote: I have used Power mean and Jensen Give your solution

\[(a^2+ab+ac+b^3/a+c^3/a)(3a+b+c) \ge 5 \sqrt[5]{a^2 b^4 c^4} \times 5 \sqrt[5]{a^3 b c}\]\[\implies \sum_{cyc} \frac{a}{a^2+b^3+c^3} \le \sum_{cyc} \frac{3 a + b + c}{25 abc} = \frac{1}{5 abc}\]

Oh man! I am soooo dumb! And I had lost hope of homogenising it!

Arhaan wrote:

How did you arrive at the 3rd line from the 2nd line?

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

RamK wrote:

I'm sorry that was a typo. There was supposed to be reciprocal of all the terms in the denominator.

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

I am feeling the same as Tom Holland did ok that bridge after that hologram thing!

Any guesses of the cut-off?

Just curious, can this be done by Denominator Nullification?

By Jensen , I did it in 5 steps

Kgxtixigct wrote: By Jensen , I did it in 5 steps How did you do that please post...

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.

Can anyone tell which function to consider to finish this problem using Jensen's inequality??? Is there any solution purely by Jensen?

$$\frac{a}{a^2+b^3+c^3}+\frac{b}{b^2+a^3+c^3}+\frac{c}{c^2+ a^3+b^3}\leq\frac{1}{5abc}$$

Let $a,b,c$ be positive real numbers such that $a+b+c=1$. Prove that $$\frac{a}{a^2+b+c}+\frac{b}{b^2+c+a}+\frac{c}{c^2+a+b}\leq\frac{1}{21abc}$$

Let $a,b,c$ be positive real numbers such that $a+b+c=1$. Prove that $$\frac{a}{a^3+b+c}+\frac{b}{b^3+c+a}+\frac{c}{c^3+a+b}\leq\frac{1}{19abc}$$

sqing wrote: Let $a,b,c$ be positive real numbers such that $a+b+c=1$. Prove that $$\frac{a}{a+b^2+c^2}+\frac{b}{b+a^2+c^2}+\frac{c}{c+a^2+b^2}\leq\frac{1}{15abc}.$$

Let $a,b,c$ be positive real numbers such that $a+b+c=1$. Prove that $$\frac{a^2}{a+b^2+c^2}+\frac{b^2}{b+a^2+c^2}+\frac{c^2}{c+a^2+b^2}\leq\frac{1}{45abc}$$$$\frac{a^2}{a^2+b+c}+\frac{b^2}{b^2+c+a}+\frac{c^2}{c^2+a+b}\leq\frac{1}{63abc}$$$$ \frac{a^2}{a^2+b^3+c^3}+\frac{b^2}{b^2+a^3+c^3}+\frac{c^2}{c^2+ a^3+b^3}\leq\frac{1}{15abc}$$$$ \frac{a^3}{a^2+b^3+c^3}+\frac{b^3}{b^2+a^3+c^3}+\frac{c^3}{c^2+ a^3+b^3}\leq\frac{1}{45abc}$$

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.

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.

Loved this ineq.