Contradiction Maybe? or holder

Yes, I had been playing around something about$a=3,b=c=1$, and this problem was proposed as its contradiction counterpart. I tried some smoothing by making $b=c$, which the maximum will only be reached when $bc<36$

For another side, I use calculus and found \[\frac{1}{b+3}+\frac{1}{c+3}\leq \sqrt{\frac{bc}{3bc-27}}\]holds for $bc>9$

supercarry wrote: Given positive real $a,b,c$ satisfying \[\frac{1}{\sqrt{a+1}}+\frac{3}{\sqrt{b+3}}+\frac{3}{\sqrt{c+3}}=\frac72\]Prove that $abc\leq 3$. I was asked to propose a inequality for the condition of $abc<3$ inequality since <3 looks like a heart shape, then I construct a equality and with the help of wolfram, I gave the birth of this bad-looking inequality, I’m glad to see any method besides calculus. We can show it by the Lagrange Multipliers method, but you want a solution without calculus

arqady wrote: supercarry wrote: Given positive real $a,b,c$ satisfying \[\frac{1}{\sqrt{a+1}}+\frac{3}{\sqrt{b+3}}+\frac{3}{\sqrt{c+3}}=\frac72\]Prove that $abc\leq 3$. I was asked to propose a inequality for the condition of $abc<3$ inequality since <3 looks like a heart shape, then I construct a equality and with the help of wolfram, I gave the birth of this bad-looking inequality, I’m glad to see any method besides calculus. We can show it by the Lagrange Multiplayers method, but you want a solution without calculus Sad, maybe lagrange solos could do

arqady wrote: supercarry wrote: Given positive real $a,b,c$ satisfying \[\frac{1}{\sqrt{a+1}}+\frac{3}{\sqrt{b+3}}+\frac{3}{\sqrt{c+3}}=\frac72\]Prove that $abc\leq 3$. I was asked to propose a inequality for the condition of $abc<3$ inequality since <3 looks like a heart shape, then I construct a equality and with the help of wolfram, I gave the birth of this bad-looking inequality, I’m glad to see any method besides calculus. We can show it by the Lagrange Multiplayers method, but you want a solution without calculus Thanks for replying and it is sad that this inequality with neat form can’t be solved by smoothing method. Actually, we have done this by calculus or Lagrange Multiplier, since it was proposed to new beginner for IMO, I think it could be monstrous problem for them.

supercarry wrote: Given positive real $a,b,c$ satisfying \[\frac{1}{\sqrt{a+1}}+\frac{3}{\sqrt{b+3}}+\frac{3}{\sqrt{c+3}}=\frac72\]Prove that $abc\leq 3$. . $\bullet \frac{5}{2} \le \frac{7}{2}-\frac{1}{\sqrt{a+1}}=\frac{3}{\sqrt{b+3}}+\frac{3}{\sqrt{c+3}} \le 3\sqrt{\frac{2}{b+3}+\frac{2}{c+3}} \Longrightarrow \frac{1}{b+3}+\frac{1}{c+3} \ge \frac{25}{72}$ $$\Longrightarrow 207 \ge 25bc+3(b+c) \ge 3bc +6\sqrt{bc} \Longrightarrow \sqrt{bc} \le \frac{69}{25} <3 \Longrightarrow bc<9. $$ $\bullet Use : \frac{1}{\sqrt{x^2+1}}+\frac{1}{\sqrt{y^2+1}} \le \frac{2}{\sqrt{xy+1}},\left(x,y \ge 0,xy\le1\right)\Longrightarrow \frac{1}{\sqrt{b+3}}+\frac{1}{\sqrt{c+3}} \le \frac{2}{\sqrt{\sqrt{bc}+3}},(1) $ $\bullet \frac{1}{\sqrt{a+1}} \le \frac{2}{\sqrt{3a}+1},(2)$ $\bullet (1),(2) \Longrightarrow \frac{7}{2}=\frac{1}{\sqrt{a+1}}+\frac{3}{\sqrt{b+3}}+\frac{3}{\sqrt{c+3}} \le \frac{2}{\sqrt{3a}+1}+\frac{6}{\sqrt{\sqrt{bc}+3}} \Longrightarrow \sqrt{bc} \le \left(\frac{12\sqrt{3a}+12}{7\sqrt{3a}+3}\right)^2-3,(3)$ $\bullet $ We need prove $ : \left(\frac{12\sqrt{3a}+12}{7\sqrt{3a}+3}\right)^2-3 \le \sqrt{\frac{3}{a}},(4) \iff \frac{12\sqrt{3a}+12}{7\sqrt{3a}+3} \le \sqrt{3+\sqrt{\frac{3}{a}}}$ $$\iff 4\left(3+\sqrt{\frac{3}{a}} \right) \le \left(7+\sqrt{\frac{3}{a}} \right)\sqrt{3+\sqrt{\frac{3}{a}}}\iff 4\sqrt{3+\sqrt{\frac{3}{a}}} \le 7+\sqrt{\frac{3}{a}} \iff \left(2-\sqrt{3+\sqrt{\frac{3}{a}}}\right)^2 \ge 0.\left(True \right)$$ $\bullet (3),(4) \Longrightarrow \sqrt{bc} \le \sqrt{\frac{3}{a}} \Longrightarrow abc \le 3.$ .

arqady wrote: supercarry wrote: Given positive real $a,b,c$ satisfying \[\frac{1}{\sqrt{a+1}}+\frac{3}{\sqrt{b+3}}+\frac{3}{\sqrt{c+3}}=\frac72\]Prove that $abc\leq 3$. I was asked to propose a inequality for the condition of $abc<3$ inequality since <3 looks like a heart shape, then I construct a equality and with the help of wolfram, I gave the birth of this bad-looking inequality, I’m glad to see any method besides calculus. We can show it by the Lagrange Multipliers method, but you want a solution without calculus The partial derivatives system may be hardcore

Drago119 wrote: The partial derivatives system may be hardcore I am ready to show my solution.

arqady wrote: Drago119 wrote: The partial derivatives system may be hardcore I am ready to show my solution. It will be great if you would like to share it with us!

supercarry wrote: Given positive real $a,b,c$ satisfying \[\frac{1}{\sqrt{a+1}}+\frac{3}{\sqrt{b+3}}+\frac{3}{\sqrt{c+3}}=\frac72\]Prove that $abc\leq 3$. Let $abc>3$, $a=3kx$, $b=ky$ and $c=kz$, where $k>0$ and $xyz=1$. Thus, $k>1$ and $$\frac{7}{2}=\frac{1}{\sqrt{3kx+1}}+\frac{3}{\sqrt{ky+3}}+\frac{3}{\sqrt{kz+3}}<\frac{1}{\sqrt{3x+1}}+\frac{3}{\sqrt{y+3}}+\frac{3}{\sqrt{z+3}},$$which is a contradiction because we'll prove now that $$\frac{1}{\sqrt{3x+1}}+\frac{3}{\sqrt{y+3}}+\frac{3}{\sqrt{z+3}}\leq\frac{7}{2}.$$Indeed, let $f(x,y,z,\lambda)=\frac{1}{\sqrt{3x+1}}+\frac{3}{\sqrt{y+3}}+\frac{3}{\sqrt{z+3}}+\lambda(xyz-1).$ If $x\rightarrow+\infty$ so $$\frac{1}{\sqrt{3x+1}}+\frac{3}{\sqrt{y+3}}+\frac{3}{\sqrt{z+3}}<2\sqrt3<\frac{7}{2}.$$If $y\rightarrow+\infty$ so $$\frac{1}{\sqrt{3x+1}}+\frac{3}{\sqrt{y+3}}+\frac{3}{\sqrt{z+3}}<1+\sqrt3<\frac{7}{2}.$$Id est, it's enough to find a maximum of $f$ in the inside point $(x,y,z)$ of some compact. But in this point we have $$\frac{\partial f}{\partial x}=\frac{\partial f}{\partial y}=\frac{\partial f}{\partial z}=0,$$which gives $$-\frac{3}{2\sqrt{(3x+1)^3}}+\lambda yz=-\frac{3}{2\sqrt{(y+3)^3}}+\lambda xz=-\frac{3}{2\sqrt{(z+3)^3}}+\lambda xy=0,$$which gives $$\frac{x}{\sqrt{(3x+1)^3}}=\frac{y}{\sqrt{(y+3)^3}}=\frac{z}{\sqrt{(z+3)^3}}$$and from the second we obtain: $$(y-z)(y^2z^2-27yz-27(y+z))=0.$$Let $y^2z^2-27yz-27(y+z)=0.$ Thus, by AM-GM $$y^2z^2=27yz+27(y+z)\geq27yz+54\sqrt{yz},$$which gives $yz\geq36$ and by Jensen: $$\frac{1}{\sqrt{3x+1}}+\frac{3}{\sqrt{y+3}}+\frac{3}{\sqrt{z+3}}<1+3\sqrt{2\left(\frac{1}{y+3}+\frac{1}{z+3}\right)}=$$$$=1+3\sqrt{\frac{2(y+z+6)}{yz+3(y+z)+9}}\leq1+3\sqrt{\frac{2(y+z+6)}{3(y+z+15)}}<1+3\sqrt{\frac{2}{3}}<\frac{7}{2}.$$Id est, it's enough to understand, what happens for $y=z$ and we need to prove that: $$\frac{y}{\sqrt{3+y^2}}+\frac{6}{\sqrt{y+3}}\leq\frac{7}{2},$$which is true by C-S and AM-GM:$$\frac{y}{\sqrt{3+y^2}}+\frac{6}{\sqrt{y+3}}=\frac{2y}{\sqrt{(3+1)(3+y^2)}}+\frac{3\sqrt{4(y+3)}}{y+3}\leq\frac{2y+\frac{3}{2}(4+y+3)}{y+3}=\frac{7}{2}.$$