Prove that for positive reals $a$,$b$,$c$ so that $a+b+c+abc=4$, \[\left (1+\dfrac{a}{b}+ca \right )\left (1+\dfrac{b}{c}+ab \right)\left (1+\dfrac{c}{a}+bc \right) \ge 27\] holds.
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Tags: inequalities
16.11.2014 22:06
$[u][b]First step[/b][/u]$ We prove that $ abc \le 1 .$ Suppose the contrary, that $ abc > 1. $ It results that $ a+b+c \ge 3\sqrt[3]{abc} > 3 \Rightarrow a + b + c + abc > 4 \Rightarrow 4 > 4$, impossible. Hence, $abc \le 1.$ Second step The inequality is equivalent with: $\left( \frac {b+a+ca} {b}\right) \left( \frac{c+b+ac}{c}\right) \left( \frac{a+c+bc}{a}\right) \ge 27 \Leftrightarrow$ $(4-a)(4-b)(4-c) \ge 27abc \Leftrightarrow 64 - 16(a+b+c) + 4(ab+bc+ca) -abc \ge 27abc \Leftrightarrow $ $64 + 4(ab+bc+ca) \ge 16(a+b+c+abc) + 12abc \Leftrightarrow 64 + 4(ab+bc+ca) \ge 64 + 12abc \Leftrightarrow $ $ab+bc+ca \ge 3abc$ Now, $ab+bc+ca \ge 3\sqrt[3]{a^2b^2c^2}$ and $\sqrt[3]{a^2b^2c^2} \ge abc \Leftrightarrow a^2b^2c^2 \ge a^3b^3c^3 \Leftrightarrow$ $1 \ge abc$ which was proven at First step
16.11.2014 22:10
bcp123 wrote: Prove that for positive reals $a$,$b$,$c$ so that $a+b+c+abc=4$, \[\left (1+\dfrac{a}{b}+ca \right )\left (1+\dfrac{b}{c}+ab \right)\left (1+\dfrac{c}{a}+bc \right) \ge 27\] holds. hello, the left side of your inequality is equivalent to $\frac{(abc+a+b)(abc+b+c)(abc+a+c)}{abc}=\frac{(4-a)(4-b)(4-c)}{abc}\geq 27$ this is equivalent to $\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\geq 3$ by AM-GM we get $\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\geq \sqrt[3]{\frac{1}{abc}}\geq 3$ if $abc\le 1$ but we have $\frac{a+b+c+abc}{4}\geq \sqrt[4]{(abc)^2}$ this is equivalent to $1\geq abc$ Sonnhard.
17.11.2014 04:38
moldovan wrote: The inequality is equivalent with:$(4-a)(4-b)(4-c) \ge 27abc .$ Generalization
24.03.2015 08:40
bcp123 wrote: Prove that for positive reals $a$,$b$,$c$ so that $a+b+c+abc=4$, \[\left (1+\dfrac{a}{b}+ca \right )\left (1+\dfrac{b}{c}+ab \right)\left (1+\dfrac{c}{a}+bc \right) \ge 27\] holds. $4=a+b+c+abc\geq4\sqrt[4]{(abc)^2} \implies abc\le 1$, $\left (1+\dfrac{a}{b}+ca \right )\left (1+\dfrac{b}{c}+ab \right)\left (1+\dfrac{c}{a}+bc \right) =\frac{(4-a)(4-b)(4-c)}{abc}$ $=15+4(\frac{1}{a}+\frac{1}{b}+\frac{1}{c})\geq 15+12\sqrt[3]{\frac{1}{abc}}\geq 27$ .
10.09.2015 03:21
First Round Pakistan (for IMO) 2016
17.12.2020 01:10
Notice that $$ a+b+c+abc=4\geq4\sqrt[4]{(abc)^2} \iff 1\geq abc \implies \frac{3}{\sqrt[3]{abc}}\geq 3 $$From the GM-HM inequality we have $$ \sqrt[3]{abc}\geq \frac{3}{\frac{1}{a}+\frac{1}{b}+\frac{1}{c}} \iff \frac{1}{a}+\frac{1}{b}+\frac{1}{c} \geq \frac{3}{\sqrt[3]{abc}}\geq 3 $$The desired inequality is $$ (a+b+abc)(b+c+abc)(c+a+abc)\geq27abc $$from the given condition this is equivalent to $$ (4-a)(4-b)(4-c)\geq 27abc $$$$ 64-16(a+b+c)+4(ab+bc+ca)\geq 28abc $$$$ 64-16(4-abc)+4(ab+bc+ca)\geq 28abc $$$$ ab+bc+ca\geq 3abc $$$$ \frac{1}{a}+\frac{1}{b}+\frac{1}{c} \geq 3 $$The latter is true from one of the aforementioned inequalities.
17.12.2022 21:02
Not bad problem : It is easy to see $\left (1+\dfrac{a}{b}+ca \right )\left (1+\dfrac{b}{c}+ab \right)\left (1+\dfrac{c}{a}+bc \right) = (a+b+c+abc)(a+b+c+abc+\frac{1}{a}+\frac{1}{b}+\frac{1}{c}) -1$ Now we need to prove $\frac{1}{a} +\frac{1}{b}+\frac{1}{c} \geq 3$ . ( ) Also we can see $a+b+c+abc = 4 \geq 4\sqrt[4]{a^2b^2c^2} \to 1\geq abc$ . ( ) By ( ) proving of ( ) is easy
24.05.2023 00:08
$1+\dfrac{a}{b}+ca=\dfrac{a+b+abc}{b}=\dfrac{4-c}{b}$ $1+\dfrac{b}{c}+ab=\dfrac{b+c+abc}{c}=\dfrac{4-a}{b}$ $1+\dfrac{c}{a}+bc=\dfrac{a+c+abc}{a}=\dfrac{4-b}{a}$ $\left (1+\dfrac{a}{b}+ca \right )\left (1+\dfrac{b}{c}+ab \right)\left (1+\dfrac{c}{a}+bc \right)=$ $\left (\dfrac{4-c}{b} \right)\left (\dfrac{4-a}{b} \right)\left (\dfrac{4-b}{a} \right) =$ $\frac{(4-a)(4-b)(4-c)}{abc}=$ $\frac{-16a-16b-16c+4ab+4bc+4ac-abc+64}{abc}=$ $\frac{64+4ab+4bc+4ac-16a-16b-16c-abc}{abc}=$ $\frac{64+4(ab+bc+ac)-16(a+b+c)-abc}{abc}$ $=>$ $\left (1+\dfrac{a}{b}+ca \right )\left (1+\dfrac{b}{c}+ab \right)\left (1+\dfrac{c}{a}+bc \right)=$ $\frac{64+4(ab+bc+ac)-16(a+b+c)-abc}{abc}$ By $AM-GM$ we get $a+b+c+abc\geq4\sqrt[4]{a*b*c*abc}$ $4\geq4\sqrt[4]{(abc)^2}$ $4\geq4\sqrt{abc}$ $1\geq\sqrt{abc}$ $1\geq abc$ $...(1)$ $ab+bc+ac\geq3\sqrt[3]{ab*bc*ac}$ $ab+bc+ac\geq3\sqrt[3]{(abc)^2}=3\sqrt[3]{(abc)^2*1}$ Combining this with $(1)$ we get: $ab+bc+ac\geq3\sqrt[3]{(abc)^2*1}\geq3\sqrt[3]{(abc)^2*abc}=3\sqrt[3]{(abc)^3}=3abc$ $=>$ $ab+bc+ac\geq3abc$ $...(2)$ $a+b+c\geq3\sqrt[3]{abc}=3\sqrt[3]{abc*1^2}$ Combining also this one with $(1)$ we get: $a+b+c\geq3\sqrt[3]{abc}=3\sqrt[3]{abc*1^2}\geq3\sqrt[3]{abc*(abc)^2}=3\sqrt[3]{abc)^3}=3abc$ $=>$ $a+b+c\geq3\sqrt[3]{abc}\geq3abc$ $...(3)$ By $(1)$,$(2)$,$(3)$ we get $\left (1+\dfrac{a}{b}+ca \right )\left (1+\dfrac{b}{c}+ab \right)\left (1+\dfrac{c}{a}+bc \right)=$ $\frac{64+4(ab+bc+ac)-16(a+b+c)-abc}{abc}\geq\frac{64abc+4*3abc-16*3abc-abc}{abc}=\frac{64abc+12abc-48abc-abc}{abc}$ $\frac{27abc}{abc}=27$ $\left (1+\dfrac{a}{b}+ca \right )\left (1+\dfrac{b}{c}+ab \right)\left (1+\dfrac{c}{a}+bc \right)\geq27$