Good inequality..

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Good inequality..

Post Number:#1  Unread postby Katy729 » Sun Jun 18, 2017 10:36 pm

Let a, b, c be real positive numbers. Prove that

$$(1+a/b)(1+b/c)(1+c/a)>=2(1+(a+b+c)/(\sqrt[3]{abc}))$$
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Re: Good inequality..

Post Number:#2  Unread postby ahmedittihad » Mon Jun 26, 2017 1:51 am

By AM-GM inequality, $ac+bc+ab \geq 3(abc)^{2/3}$.
Also by AM-GM inequality, $a+b+c \geq 3(abc)^{1/3}$.
So, $3 \geq 2+ \frac{3 (abc)^{1/3}}{(a+b+c)}$.
Or, $ac+bc+ab \geq 3(abc)^{2/3} \geq (abc)^{2/3} (2+ \frac{3 (abc)^{1/3}}{(a+b+c)})$.
Or, $ac+bc+ab \geq 2(abc)^{2/3} + \frac {3abc}{a+b+c}$.
Or, $ac+bc+ab- \frac {3abc}{a+b+c} \geq 2(abc)^{2/3}$.
Or, $\frac {ac^2+a^2c+bc^2+b^2c+ab^2+a^2b}{a+b+c} \geq 2(abc)^{2/3}$.
Or, $\frac {ac^2+a^2c+bc^2+b^2c+ab^2+a^2b}{a+b+c} \geq \frac {2abc}{(abc)^{1/3}}$.
Or, $\frac {ac^2+a^2c+bc^2+b^2c+ab^2+a^2b}{abc} \geq \frac {2(a+b+c)}{(abc)^{1/3}}$
Or, $\frac {a}{b}+\frac {b}{a}+\frac {a}{c}+\frac {c}{a}+\frac {b}{c}+\frac {c}{b}\geq \frac {2(a+b+c)}{(abc)^{1/3}}$
Or, $2+\frac {a}{b}+\frac {b}{a}+\frac {a}{c}+\frac {c}{a}+\frac {b}{c}+\frac {c}{b}\geq 2+ \frac {2(a+b+c)}{(abc)^{1/3}}$.
Or, $(1+\frac {a}{b})(1+\frac {b}{c})(1+\frac {c}{a})\geq 2(1+\frac {a+b+c}{(abc)^{1/3}})$.
Which is indeed our problem statement.
I recommend you use \frac function to latex fractions. It's prettier and easier to interpret.
Frankly, my dear, I don't give a damn.
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Re: Good inequality..

Post Number:#3  Unread postby Katy729 » Sat Jul 01, 2017 3:46 pm

good and clear solution. Thanks! :)
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