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Question Number 84420 by Power last updated on 12/Mar/20

Commented by Power last updated on 12/Mar/20

$$\mathrm{prove}\mathrm{it}$$

Commented by Power last updated on 12/Mar/20

$$\frac{\mathrm{a}}{\mathrm{b}}+\frac{\mathrm{b}}{\mathrm{c}}+\frac{\mathrm{c}}{\mathrm{a}}+3⩾\sqrt{\mathrm{a}}+\sqrt{\mathrm{b}}+\sqrt{\mathrm{c}}$$

Answered by mind is power last updated on 12/Mar/20

$$\frac{a}{b}+\frac{b}{c}+\frac{c}{a}+3⩾\sqrt{a}+\sqrt{b}+\sqrt{c}$$$$\frac{a}{b}+\frac{b}{c}+\frac{c}{a}+\frac{1}{4}(a+b+c)=\frac{a}{b}+\frac{b}{4}+\frac{b}{c}+\frac{c}{4}+\frac{c}{a}+\frac{a}{4}$$$$AM-GM\Rightarrow \frac{a}{b}+\frac{b}{4}⩾2\sqrt{\frac{a}{b}.\frac{b}{4}}=\sqrt{a}$$$$\frac{b}{c}+\frac{c}{4}⩾\sqrt{b},\frac{c}{a}+\frac{a}{4}⩾\sqrt{c}\Rightarrow$$$$\frac{a}{b}+\frac{b}{c}+\frac{c}{a}+\frac{a+b+c}{4}⩾\sqrt{a}+\sqrt{b}+\sqrt{c}$$$$\iff \frac{a}{b}+\frac{b}{c}+\frac{c}{a}+3⩾\sqrt{a}+\sqrt{b}+\sqrt{c}$$$$conditionis(a,b,c)\in {\mathbb{R}}_{+}^3\mid a+b+c=12not\in \mathbb{N}only$$

Commented by Power last updated on 13/Mar/20

$$\mathrm{thanks}$$

Commented by mind is power last updated on 13/Mar/20

$$pleasur$$

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