# 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!

Note by Figel Ilham
3 years, 1 month ago

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

Staff - 3 years, 1 month ago

Does the proof true? Consider $$\frac{a^2}{b}+\frac{a^2}{b} +\frac{b^2}{c}+c$$ Applying AM-GM, we have $\frac{a^2}{b} +\frac{a^2}{b} +\frac{b^2}{c}+c \geq 4\sqrt[4]{\frac{a^2}{b} \frac{a^2}{b} \frac{b^2}{c} c} = 4a$ $2\frac{a^2}{b} +\frac{b}{c} +c \geq 4a$ Apply also to $2\frac{b^2}{c} +\frac{c}{d} +d \geq 4b$ $2\frac{c^2}{d} +\frac{d^2}{a} +a \geq 4c$ $2\frac{d^2}{a} +\frac{a^2}{b} +b \geq 4d$

Adding those, we have $3(\frac{a^2}{b} + \frac{b^2}{c} +\frac{c^2}{d} +\frac{d^2}{a}) +(a+b+c+d) \geq 4(a+b+c+d)$ $3(\frac{a^2}{b} + \frac{b^2}{c} +\frac{c^2}{d} +\frac{d^2}{a}) \geq 3(a+b+c+d)$ $\frac{a^2}{b} + \frac{b^2}{c} +\frac{c^2}{d} +\frac{d^2}{a} ) \geq (a+b+c+d)$

- 3 years, 1 month ago

Perfect! Well done :)

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

Staff - 3 years, 1 month ago

I'll try

- 3 years, 1 month ago

I'll try first

- 3 years, 1 month ago