Let $a,b,c$ be non-negative numbers such that $a+b+c=3.$ Prove that \[\sqrt{3a^3+4bc+b+c}+\sqrt{3b^3+4ca+c+a}+\sqrt{3c^3+4ab+a+b} \geqslant 9.\]
Problem
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Tags: inequalities
Matricy
27.12.2024 04:46
Lagrange multipliers.
arqady
27.12.2024 04:49
Matricy wrote: Lagrange multipliers. Does not work. Holder with $(a+2)^3$ helps, but it's very ugly.
samrocksnature
27.12.2024 04:56
Beautiful problem.
hamir
27.12.2024 05:01
@samrocksnature agreed.
YaoAOPS
27.12.2024 05:03
I've enjoyed every minute I've spent thinking about this problem.
MyLifeMyChoice
27.12.2024 07:26
arqady wrote: Matricy wrote: Lagrange multipliers. Does not work. Holder with $(a+2)^3$ helps, but it's very ugly. I think we could use CBS first to simplify the problem: $\sqrt{3a^3+4bc+b+c}\ge\frac{3a^2+4bc+b+c}{\sqrt{3a+4bc+b+c}}$ and its analogs, then further applying Holder
arqady
27.12.2024 13:36
MyLifeMyChoice wrote: I think we could use CBS first to simplify the problem: $\sqrt{3a^3+4bc+b+c}\ge\frac{3a^2+4bc+b+c}{\sqrt{3a+4bc+b+c}}$ and its analogs, then further applying Holder No! Now use AM-GM and SOS. Turned out nice!
puntre
04.01.2025 13:20
Very neat problem. Here is my solution (This solution is wrong, $A$, $B$, $C$ have to be adjust a little bit more like in #13) Link to pdf This problem is by far one of the nicest IE I have solved (because even if the solution looks like a standard inequality problem, what was going on behind the scene is largely based on combinatorics). During my solving process, one key soft technique that I used is to look at a bigger picture $\sqrt P+\sqrt Q+\sqrt R$ for some polynomials $P$, $Q$, and $R$. I felt like there should be a nice bound for this summation, at which point this problem seems more like a combinatorial optimization problem. Thanks @below for pointing out
arqady
05.01.2025 04:10
puntre wrote: Link to pdf I think, the statement $A+B+C=X+Y+Z$ is wrong.
ItzsleepyXD
05.01.2025 18:19
I am so sad that tkn didn't write solution for me and I have to type latex for 1 hour and it took so long to solve (7 hour)
WLOG $a \geqslant b \geqslant c$ so $ 3 \geqslant a \geqslant 1 , ab \geqslant ac \geqslant bc$
Let $$ X = {3a^3+4bc+b+c} , Y = 3b^3+4ca+c+a , Z = 3c^3+4ab+a+b$$$$ P = 3a^3-2a^2+5a+3 , Q = 3b^3-2b^2+5a+3 , R = 3c^3-2c^2+6c+3 $$Known that $ P = 3a^3-2a^2+5a+3 = 3a^3+2ac+2ab+b+c$
It is easy to see that $$X+Y+Z=P+Q+R$$and want to prove that $$\sqrt{X}+\sqrt{Y}+\sqrt{Z} \geqslant \sqrt{P} + \sqrt{R} + \sqrt{Z}$$1. Karamata$\textcolor{red}{\boxed{ \text{Claim 1 } P \geqslant X}}$
Consider $ab+ac \geqslant 2bc$
so $3a^3+2ac+2ab+b+c \geqslant 3a^3+4bc+b+c$
$\textcolor{red}{\boxed{ \text{Claim 2 } P+Q \geqslant X+Y}}$
Consider $2ab \geqslant ac+bc$
so $ 3a^3+2ab+2ac+b+c+3b^3+2ab+2bc+a+c \geqslant 3a^3+2bc+2bc+b+c + 3b^3 +2ac +2ac + a + c$
Then $(P,Q,R)$ majorize $(X,Y,Z)$ and known that $f(x) = \sqrt{x}$ is concave .
By Karamata $$\sqrt{X}+\sqrt{Y}+\sqrt{Z} \geqslant \sqrt{P} + \sqrt{R} + \sqrt{Z}$$
Also Want to prove that $$\sqrt{P} + \sqrt{R} + \sqrt{Z} \geqslant 9$$2. Tangent Line Trick (Inspire From Puntre LOL$$\textcolor{blue}{\boxed{g(x) \geqslant g(a) + (x-a)g'(a)}}$$consider $\textcolor{red}{g(x)=\sqrt{3x^3-2x^2+5x+3}}$
choose $a=1$ then $g(1)=3,g'(1)= \frac{5}{3}$
so $\textcolor{magenta}{g(x) \geqslant 3+ \frac{5(x-1)}{3}}$
then
$$\boxed{\textcolor{blue}{\sqrt{P} + \sqrt{R} + \sqrt{Z} \geqslant 9}}$$
So
$$\boxed{\textcolor{magenta}{\sqrt{3a^3+4bc+b+c}+\sqrt{3b^3+4ca+c+a}+\sqrt{3c^3+4ab+a+b} \geqslant 9}}$$$\boxed{\textcolor{blue}{\text{Finished }} \textcolor{red}{\square}}$
truongphatt2668
19.01.2025 09:19
ItzsleepyXD wrote:
I am so sad that tkn didn't write solution for me and I have to type latex for 1 hour and it took so long to solve (7 hour)
WLOG $a \geqslant b \geqslant c$ so $ 3 \geqslant a \geqslant 1 , ab \geqslant ac \geqslant bc$
Let $$ X = {3a^3+4bc+b+c} , Y = 3b^3+4ca+c+a , Z = 3c^3+4ab+a+b$$$$ P = 3a^3-2a^2+5a+3 , Q = 3b^3-2b^2+5a+3 , R = 3c^3-2c^2+6c+3 $$Known that $ P = 3a^3-2a^2+5a+3 = 3a^3+2ac+2ab+b+c$
It is easy to see that $$X+Y+Z=P+Q+R$$and want to prove that $$\sqrt{X}+\sqrt{Y}+\sqrt{Z} \geqslant \sqrt{P} + \sqrt{R} + \sqrt{Z}$$1. Karamata$\textcolor{red}{\boxed{ \text{Claim 1 } P \geqslant X}}$
Consider $ab+ac \geqslant 2bc$
so $3a^3+2ac+2ab+b+c \geqslant 3a^3+4bc+b+c$
$\textcolor{red}{\boxed{ \text{Claim 2 } P+Q \geqslant X+Y}}$
Consider $2ab \geqslant ac+bc$
so $ 3a^3+2ab+2ac+b+c+3b^3+2ab+2bc+a+c \geqslant 3a^3+2bc+2bc+b+c + 3b^3 +2ac +2ac + a + c$
Then $(P,Q,R)$ majorize $(X,Y,Z)$ and known that $f(x) = \sqrt{x}$ is concave .
By Karamata $$\sqrt{X}+\sqrt{Y}+\sqrt{Z} \geqslant \sqrt{P} + \sqrt{R} + \sqrt{Z}$$
Also Want to prove that $$\sqrt{P} + \sqrt{R} + \sqrt{Z} \geqslant 9$$2. Tangent Line Trick (Inspire From Puntre LOL$$\textcolor{blue}{\boxed{g(x) \geqslant g(a) + (x-a)g'(a)}}$$consider $\textcolor{red}{g(x)=\sqrt{3x^3-2x^2+5x+3}}$
choose $a=1$ then $g(1)=3,g'(1)= \frac{5}{3}$
so $\textcolor{magenta}{g(x) \geqslant 3+ \frac{5(x-1)}{3}}$
then
$$\boxed{\textcolor{blue}{\sqrt{P} + \sqrt{R} + \sqrt{Z} \geqslant 9}}$$
So
$$\boxed{\textcolor{magenta}{\sqrt{3a^3+4bc+b+c}+\sqrt{3b^3+4ca+c+a}+\sqrt{3c^3+4ab+a+b} \geqslant 9}}$$$\boxed{\textcolor{blue}{\text{Finished }} \textcolor{red}{\square}}$
How can you choose $X,Y,Z,P,Q,R$ like that? Thank a lots