"Determine, with certainty"? Hmmm... We claim that the maximum is $1$, attained when $a=b=c=1$. Indeed, we have $\sum_{cyc} ab\le \frac13(a+b+c)^2=3$, Also, Cauchy-Schwarz gives $$\left(a^2+\frac{b^2}{a}+\frac{c}{a}\right)(1+a+ca)\ge (a+b+c)^2\implies \frac{1}{a^2+\frac{b^2}{a}+\frac{c}{a}}\le \frac{1+a+ca}{(a+b+c)^2}$$Using this and similar cyclic results, $$\frac{a}{a^3+b^2+c}+\frac{b}{b^3+c^2+a}+\frac{c}{c^3+a^2+b}= \sum _{cyc} \frac{1}{a^2+\frac{b^2}{a}+\frac{c}{a}}\le \sum_{cyc} \frac{1+a+ca}{(a+b+c)^2}=\frac19\left( 3+3+\sum_{cyc} ab\right)\le \frac19(3+3+3)=1.$$Hence the result.

@above, Cauchy-Schwarz all the way! Nice! Mine was quite gory, upon homogenising and then using SOS. Really nice solution, again.

anantmudgal09 wrote: Let $a,b,c$ be positive real numbers such that $a+b+c=3$. Determine, with certainty, the largest possible value of the expression $$ \frac{a}{a^3+b^2+c}+\frac{b}{b^3+c^2+a}+\frac{c}{c^3+a^2+b}$$ $$ \frac{a}{a^3+b^2+c}+\frac{b}{b^3+c^2+a}+\frac{c}{c^3+a^2+b}\le \frac{a}{2a+b}+\frac{b}{2b+c}+\frac{c}{2c+a}\le 1$$See also here

sqing wrote: anantmudgal09 wrote: Let $a,b,c$ be positive real numbers such that $a+b+c=3$. Determine, with certainty, the largest possible value of the expression $$ \frac{a}{a^3+b^2+c}+\frac{b}{b^3+c^2+a}+\frac{c}{c^3+a^2+b}$$ $$ \frac{a}{a^3+b^2+c}+\frac{b}{b^3+c^2+a}+\frac{c}{c^3+a^2+b}\le \frac{a}{2a+b}+\frac{b}{2b+c}+\frac{c}{2c+a}\le 1$$See also here $\frac{a}{a^3+b^2+c} \le (AM-GM)\frac{a}{3a+2b+c-3}=\frac{a}{2a+b}$. So $\frac{a}{a^3+b^2+c}+\frac{b}{b^3+c^2+a}+\frac{c}{c^3+a^2+b}\le \frac{a}{2a+b}+\frac{b}{2b+c}+\frac{c}{2c+a}$. Now, $\frac{a}{2a+b}+\frac{b}{2b+c}+\frac{c}{2c+a}\le \ 1 \Leftrightarrow \frac{b}{2a+b}+\frac{c}{2b+c}+\frac{a}{2c+a} \ge \ 1 $. Finally, $ \sum{\frac{b}{2a+b}}=\sum{\frac{b^2}{2ab+b^2}} \ge \frac{(a+b+c)^2}{a^2+b^2+c^2+2ab+2ac+2bc}=1$ .Done.

anantmudgal09 wrote: Let $a,b,c$ be positive real numbers such that $a+b+c=3$. Determine, with certainty, the largest possible value of the expression $$ \frac{a}{a^3+b^2+c}+\frac{b}{b^3+c^2+a}+\frac{c}{c^3+a^2+b}$$ 2013 Beijing University Science camps Q2:

here

My solution: Let the given expression be $E$. We claim that the maximum value of $E$ is $1$. This is attained when $a=b=c=1$. In what follows, $\sum$ shall denote cyclic summation over $a,b,c$. Notice that $$a^3-(3a-2)=(a-1)^2(a+2) \ge 0$$and $$b^2-(2b-1)=(b-1)^2 \ge 0.$$We obtain that $$\sum \frac{a}{a^3+b^2+c} \le \sum \frac{a}{3a+2b+c-3}=\sum \frac{a}{2a+b}=\frac{1}{2}\sum \left(1-\frac{b}{2a+b}\right)$$and as $$\sum \frac{b}{2a+b}=\sum \frac{b^2}{b^2+2ab} \ge \frac{(a+b+c)^2}{\sum (b^2+2ab)}=1$$by the Cauchy-Schwartz inequality, it follows that $E \le 1$.

Use am hm $P(a,b,c)\le 1$ Since $\sum_{\text{cyc}}\frac{a^2}{b}\ge \frac{(a+b+c)^2}{a+b+c}=3$ And $\sum_{\text{cyc}} \frac{a}{b}\ge 3$ $a^2+b^2+c^2=9–2(ab+bc+ac)\ge 9–2(a^2+b^2+c^2)$ thus $a^2+b^2+c^2\ge 3$ Just put all this in $AMof (–1)th power\ge (–1)th power of AM$ to get max of $1$

sqing wrote: anantmudgal09 wrote: Let $a,b,c$ be positive real numbers such that $a+b+c=3$. Determine, with certainty, the largest possible value of the expression $$ \frac{a}{a^3+b^2+c}+\frac{b}{b^3+c^2+a}+\frac{c}{c^3+a^2+b}$$ 2013 Beijing University Science camps Q2:

here Let $a,b,c$ be positive real numbers such that $ab+bc+ca=3$. Prove that$$ \frac{a}{a^3+b^2+c}+\frac{b}{b^3+c^2+a}+\frac{c}{c^3+a^2+b}\leq 1.$$Let $a,b,c$ be positive real numbers such that $ab+bc+ca=3$. Prove that$$ \frac{1}{a^3+b^2+c}+\frac{1}{b^3+c^2+a}+\frac{1}{c^3+a^2+b}\leq 1.$$

sqing wrote: sqing wrote: anantmudgal09 wrote: Let $a,b,c$ be positive real numbers such that $a+b+c=3$. Determine, with certainty, the largest possible value of the expression $$ \frac{a}{a^3+b^2+c}+\frac{b}{b^3+c^2+a}+\frac{c}{c^3+a^2+b}$$ 2013 Beijing University Science camps Q2:

here Let $a,b,c$ be positive real numbers such that $ab+bc+ca=3$. Prove that$$ \frac{a}{a^3+b^2+c}+\frac{b}{b^3+c^2+a}+\frac{c}{c^3+a^2+b}\leq 1.$$Let $a,b,c$ be positive real numbers such that $ab+bc+ca=3$. Prove that$$ \frac{1}{a^3+b^2+c}+\frac{1}{b^3+c^2+a}+\frac{1}{c^3+a^2+b}\leq 1.$$ Pls give some clarity on this

Ankoganit wrote: "Determine, with certainty"? Hmmm... We claim that the maximum is $1$, attained when $a=b=c=1$. Indeed, we have $\sum_{cyc} ab\le \frac13(a+b+c)^2=3$, Also, Cauchy-Schwarz gives $$\left(a^2+\frac{b^2}{a}+\frac{c}{a}\right)(1+a+ca)\ge (a+b+c)^2\implies \frac{1}{a^2+\frac{b^2}{a}+\frac{c}{a}}\le \frac{1+a+ca}{(a+b+c)^2}$$Using this and similar cyclic results, $$\frac{a}{a^3+b^2+c}+\frac{b}{b^3+c^2+a}+\frac{c}{c^3+a^2+b}= \sum _{cyc} \frac{1}{a^2+\frac{b^2}{a}+\frac{c}{a}}\le \sum_{cyc} \frac{1+a+ca}{(a+b+c)^2}=\frac19\left( 3+3+\sum_{cyc} ab\right)\le \frac19(3+3+3)=1.$$Hence the result. Will it not be $$\left(a^2+\frac{b^2}{a}+\frac{c}{a}\right)(1+a+ca)\ge (a^2+b^2+c^2)^2$$instead of $$\left(a^2+\frac{b^2}{a}+\frac{c}{a}\right)(1+a+ca)\ge (a+b+c)^2$$ And you cannot say that $$(a+b+c)^2>(a^2+b^2+c^2)^2$$As consider $a=2,b=0.5,c=0.5$

Ankoganit wrote: "Determine, with certainty"? Hmmm... We claim that the maximum is $1$, attained when $a=b=c=1$. Indeed, we have $\sum_{cyc} ab\le \frac13(a+b+c)^2=3$, Also, Cauchy-Schwarz gives $$\left(a^2+\frac{b^2}{a}+\frac{c}{a}\right)(1+a+ca)\ge (a+b+c)^2\implies \frac{1}{a^2+\frac{b^2}{a}+\frac{c}{a}}\le \frac{1+a+ca}{(a+b+c)^2}$$Using this and similar cyclic results, $$\frac{a}{a^3+b^2+c}+\frac{b}{b^3+c^2+a}+\frac{c}{c^3+a^2+b}= \sum _{cyc} \frac{1}{a^2+\frac{b^2}{a}+\frac{c}{a}}\le \sum_{cyc} \frac{1+a+ca}{(a+b+c)^2}=\frac19\left( 3+3+\sum_{cyc} ab\right)\le \frac19(3+3+3)=1.$$Hence the result. If I do the correction for you, it will look like- $$(a^2+b^2+c^2)^2$$$$=\bigg((a+b+c)^2-2\sum_{cyc}ab\bigg)^2$$Note that for the original inequality to be maximum, we have to take the maximum value of $\sum_{cyc}ab$ and that's what we do -- $$=\bigg(9-(2.3)\bigg)^2$$$$=(3)^2=9$$So as you can see, at the end your result matches because it indeed comes out to be $9$, but there was this mistake in your calculation.. although I must admit, your thought was beautiful nevertheless.. If you find any flaw in my reasoning, please do correct me.. I will be much humble to that

stranger_02 wrote: Ankoganit wrote: "Determine, with certainty"? Hmmm... We claim that the maximum is $1$, attained when $a=b=c=1$. Indeed, we have $\sum_{cyc} ab\le \frac13(a+b+c)^2=3$, Also, Cauchy-Schwarz gives $$\left(a^2+\frac{b^2}{a}+\frac{c}{a}\right)(1+a+ca)\ge (a+b+c)^2\implies \frac{1}{a^2+\frac{b^2}{a}+\frac{c}{a}}\le \frac{1+a+ca}{(a+b+c)^2}$$Using this and similar cyclic results, $$\frac{a}{a^3+b^2+c}+\frac{b}{b^3+c^2+a}+\frac{c}{c^3+a^2+b}= \sum _{cyc} \frac{1}{a^2+\frac{b^2}{a}+\frac{c}{a}}\le \sum_{cyc} \frac{1+a+ca}{(a+b+c)^2}=\frac19\left( 3+3+\sum_{cyc} ab\right)\le \frac19(3+3+3)=1.$$Hence the result. If I do the correction for you, it will look like- $$(a^2+b^2+c^2)^2$$$$=\bigg((a+b+c)^2-2\sum_{cyc}ab\bigg)^2$$Note that for the original inequality to be maximum, we have to take the maximum value of $\sum_{cyc}ab$ and that's what we do -- $$=\bigg(9-(2.3)\bigg)^2$$$$=(3)^2=9$$So as you can see, at the end your result matches because it indeed comes out to be $9$, but there was this mistake in your calculation.. although I must admit, your thought was beautiful nevertheless.. If you find any flaw in my reasoning, please do correct me.. I will be much humble to that Dude...you take the square roots when doing CS. Also, and this is a common mistake, for the expression to be maximised, you need the minimum, not maximum of $ab+bc+ca$ (notice the minus in front of it).

Math00954 wrote: stranger_02 wrote: Ankoganit wrote: "Determine, with certainty"? Hmmm... We claim that the maximum is $1$, attained when $a=b=c=1$. Indeed, we have $\sum_{cyc} ab\le \frac13(a+b+c)^2=3$, Also, Cauchy-Schwarz gives $$\left(a^2+\frac{b^2}{a}+\frac{c}{a}\right)(1+a+ca)\ge (a+b+c)^2\implies \frac{1}{a^2+\frac{b^2}{a}+\frac{c}{a}}\le \frac{1+a+ca}{(a+b+c)^2}$$Using this and similar cyclic results, $$\frac{a}{a^3+b^2+c}+\frac{b}{b^3+c^2+a}+\frac{c}{c^3+a^2+b}= \sum _{cyc} \frac{1}{a^2+\frac{b^2}{a}+\frac{c}{a}}\le \sum_{cyc} \frac{1+a+ca}{(a+b+c)^2}=\frac19\left( 3+3+\sum_{cyc} ab\right)\le \frac19(3+3+3)=1.$$Hence the result. If I do the correction for you, it will look like- $$(a^2+b^2+c^2)^2$$$$=\bigg((a+b+c)^2-2\sum_{cyc}ab\bigg)^2$$Note that for the original inequality to be maximum, we have to take the maximum value of $\sum_{cyc}ab$ and that's what we do -- $$=\bigg(9-(2.3)\bigg)^2$$$$=(3)^2=9$$So as you can see, at the end your result matches because it indeed comes out to be $9$, but there was this mistake in your calculation.. although I must admit, your thought was beautiful nevertheless.. If you find any flaw in my reasoning, please do correct me.. I will be much humble to that Dude...you take the square roots when doing CS. Also, and this is a common mistake, for the expression to be maximised, you need the minimum, not maximum of $ab+bc+ca$ (notice the minus in front of it). Ohh come on!!! .. you are right! I keep forgetting that .. My bad, I will delete my post okay.. and what I said about maximising was right because that expression was at the denominator, so you have to make that least in order to make the overall maximum.. to make it least, you have to choose the maximum value of $\sum_{cyc}ab$ as there's a $-ve$ sign before that as you have seen.. Wait! Why can't I delete it?

Let $a,b,c$ be positive real numbers such that $(a-2)(b-2)(c-2)+2\geq abc .$ Prove that$$ \frac{a}{a^3+b^2+c}+\frac{b}{b^3+c^2+a}+\frac{c}{c^3+a^2+b}\leq \frac{3}{a+b+c}$$h

@Ankoganit Directly apply titus lemma.Works better

We have $a,b,c \in \mathbb{R^{+}}$ such that $a+b+c=3$. Claim: $\color{blue}{\sum_{cyc} \frac{a}{a^{3}+b^{2}+c} \leq 1}$ Proof: Here, $a^{3}+b^{2}+c=a^{3}+b^{2}+(3-a-b)=(a^{3}-a+3)+(b^{2}-b)$. Now note that, $$a^{3}+1+1 \geq 3a \implies \boxed{a^{3}-a+3 \geq 2a+1}$$$$(b-1)^{2} \geq 0 \implies b^{2}-2b+1 \geq 0 \implies \boxed{b^{2}-b \geq b-1}$$Hence, $a^{3}+b^{2}+c \geq 2a+b$ $\implies$ $\boxed{\sum_{cyc} \frac{a}{a^{3}+b^{2}+c} \leq \sum_{cyc} \frac{a}{2a+b}}$ Thus we wish that: $$\sum_{cyc} \frac{a}{2a+b} \leq 1$$$$ \iff a(2b+c)(2c+a)+b(2a+b)(2c+a)+c(2a+b)(2b+c) \leq (2a+b)(2b+c)(2c+a) $$$$ \iff 12abc+4a^2b+4b^2c+4c^2a+ab^2+bc^2+ca^2 \leq 9abc+4a^2b+4b^2c+4c^2a+2ab^2+2bc^2+2ca^2$$$$ \iff 3abc \leq ab^2+bc^2+ca^2 $$which is indeed true by AM-GM inequality. $\blacksquare$ $\mathbb{(QED)}$ $\color{blue}{\text{Equality holds at}}$ $\color{blue}{\boxed{a=b=c=1}}$.

Guys what

My brain hurts :noo:

It is so similar to JBMO Shortlist 2023 #A.4 and JBMO Shortlist 2019 #A.5. Use Cauchy-Schwarz as we did in those problems $$\sum_{cyc}{\dfrac{a}{a^3+b^2+c}}\leq \dfrac{\sum\limits_{cyc}{a\left(\dfrac{1}{a}+1+c\right)}}{(a+b+c)^2}=\dfrac{ab+bc+ca+6}{9}\leq 1$$as desired.

$$\frac{a}{a^3+b^2+c}+\frac{b}{b^3+c^2+a}+\frac{c}{c^3+a^2+b} = \frac{a}{a\cdot a^2+b \cdot b+c\cdot 1}+\frac{b}{b\cdot b^2+c \cdot c+a\cdot 1}+\frac{c}{c\cdot c^2+a \cdot a+b\cdot 1} = \frac{1}{3} \left( \frac{a}{\frac{a\cdot a^2+b \cdot b+c\cdot 1}{a+b+c}}+\frac{b}{\frac{b\cdot b^2+c \cdot c+a\cdot 1}{a+b+c}}+\frac{c}{\frac{c\cdot c^2+a \cdot a+b\cdot 1}{a+b+c}} \right) \leq \frac{1}{3} \left( \frac{a}{\sqrt[3]{a^2b}} + \frac{b}{\sqrt[3]{b^2c}} + \frac{c}{\sqrt[3]{c^2a}} \right) = \frac{1}{3} \left( \sqrt[3]{\frac{a}{b}} + \sqrt[3]{\frac{b}{c}} + \sqrt[3]{\frac{c}{a}} \right) \leq \frac{1}{3} \left( (a+b+c)^{\frac{1}{3}} \left( \frac{1}{a} + \frac{1}{b} + \frac{1}{c} \right)^{\frac{1}{3}} (3)^{\frac{1}{3}} \right) \leq \frac{1}{3} (27)^{\frac{1}{3}} = 1$$