AM-GM Struggle (2)!

I have struggled on this proof. Please a hint, especially related to AM-GM (Cauchy-Schwarz, Jensen are also okay). For \(a,b,c,d \in \mathbb{R}^{+}\), proof that \[\frac{a^2}{b} +\frac{b^2}{c} +\frac{c^2}{d} +\frac{d^2}{a} \geq a+b+c+d\] Thank you very much!

#Algebra #Inequality #Struggle

Note by Figel Ilham
6 years, 1 month ago

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Comments

Check out Applying AM-GM inequality wiki, and in particular Rearranging creatively.

Calvin Lin Staff - 6 years, 1 month ago

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I'll try first

Figel Ilham - 6 years, 1 month ago

Does the proof true? Consider a2b+a2b+b2c+c\frac{a^2}{b}+\frac{a^2}{b} +\frac{b^2}{c}+c Applying AM-GM, we have a2b+a2b+b2c+c≥4a2ba2bb2cc4=4a\frac{a^2}{b} +\frac{a^2}{b} +\frac{b^2}{c}+c \geq 4\sqrt[4]{\frac{a^2}{b} \frac{a^2}{b} \frac{b^2}{c} c} = 4a 2a2b+bc+c≥4a2\frac{a^2}{b} +\frac{b}{c} +c \geq 4a Apply also to 2b2c+cd+d≥4b2\frac{b^2}{c} +\frac{c}{d} +d \geq 4b 2c2d+d2a+a≥4c2\frac{c^2}{d} +\frac{d^2}{a} +a \geq 4c 2d2a+a2b+b≥4d2\frac{d^2}{a} +\frac{a^2}{b} +b \geq 4d

Adding those, we have 3(a2b+b2c+c2d+d2a)+(a+b+c+d)≥4(a+b+c+d)3(\frac{a^2}{b} + \frac{b^2}{c} +\frac{c^2}{d} +\frac{d^2}{a}) +(a+b+c+d) \geq 4(a+b+c+d) 3(a2b+b2c+c2d+d2a)≥3(a+b+c+d)3(\frac{a^2}{b} + \frac{b^2}{c} +\frac{c^2}{d} +\frac{d^2}{a}) \geq 3(a+b+c+d) a2b+b2c+c2d+d2a)≥(a+b+c+d)\frac{a^2}{b} + \frac{b^2}{c} +\frac{c^2}{d} +\frac{d^2}{a} ) \geq (a+b+c+d)

Figel Ilham - 6 years, 1 month ago

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Perfect! Well done :)

Could you add this as an example to the wiki page?

Calvin Lin Staff - 6 years, 1 month ago

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@Calvin Lin – I'll try

Figel Ilham - 6 years, 1 month ago
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