Let $a,b,c$ be positive real numbers such that $a+b+c=3$. Determine, with certainty, the largest possible value of the expression $$ \frac{a}{a^3+b^2+c}+\frac{b}{b^3+c^2+a}+\frac{c}{c^3+a^2+b}$$
Problem
Source: RMO Delhi 2016, P4
Tags: Inequality, algebra, three variable inequality, maximum value, inequalities
11.10.2016 10:43
"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.
11.10.2016 11:21
@above, Cauchy-Schwarz all the way! Nice! Mine was quite gory, upon homogenising and then using SOS. Really nice solution, again.
11.10.2016 13:07
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
11.10.2016 15:50
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.
12.10.2016 03:15
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
14.10.2016 14:57
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 $E \le 1$.
26.04.2018 14:07
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 $1$
17.07.2018 16:46
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$$ \frac{1}{a^3+b^2+c}+\frac{1}{b^3+c^2+a}+\frac{1}{c^3+a^2+b}\leq 1.$$
18.10.2019 15:24
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
28.06.2020 23:08
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 $a=2,b=0.5,c=0.5$
28.06.2020 23:22
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
28.06.2020 23:32
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).
29.06.2020 00:42
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?
08.04.2022 15:47
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
29.10.2024 14:20
@Ankoganit Directly apply titus lemma.Works better
02.11.2024 18:28
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. $\blacksquare$ $\mathbb{(QED)}$ $\color{blue}{\text{Equality holds at}}$ $\color{blue}{\boxed{a=b=c=1}}$.
02.11.2024 18:32
Guys what
02.11.2024 18:33
My brain hurts :noo:
02.11.2024 20:24
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.
11.11.2024 21:32
$$\frac{a}{a^3+b^2+c}+\frac{b}{b^3+c^2+a}+\frac{c}{c^3+a^2+b} = \frac{a}{a\cdot a^2+b \cdot b+c\cdot 1}+\frac{b}{b\cdot b^2+c \cdot c+a\cdot 1}+\frac{c}{c\cdot c^2+a \cdot a+b\cdot 1} = \frac{1}{3} \left( \frac{a}{\frac{a\cdot a^2+b \cdot b+c\cdot 1}{a+b+c}}+\frac{b}{\frac{b\cdot b^2+c \cdot c+a\cdot 1}{a+b+c}}+\frac{c}{\frac{c\cdot c^2+a \cdot a+b\cdot 1}{a+b+c}} \right) \leq \frac{1}{3} \left( \frac{a}{\sqrt[3]{a^2b}} + \frac{b}{\sqrt[3]{b^2c}} + \frac{c}{\sqrt[3]{c^2a}} \right) = \frac{1}{3} \left( \sqrt[3]{\frac{a}{b}} + \sqrt[3]{\frac{b}{c}} + \sqrt[3]{\frac{c}{a}} \right) \leq \frac{1}{3} \left( (a+b+c)^{\frac{1}{3}} \left( \frac{1}{a} + \frac{1}{b} + \frac{1}{c} \right)^{\frac{1}{3}} (3)^{\frac{1}{3}} \right) \leq \frac{1}{3} (27)^{\frac{1}{3}} = 1$$