Prove-that-a-b-b-c-c-a-a-2-1-b-2-1-b-2-1-c-2-1-c-2-1-a-2-1-for-a-b-c-are-positive-real-number-

Question Number 150039 by bobhans last updated on 09/Aug/21

Prove that \frac{a}{b} + \frac{b}{c} + \frac{c}{a} ⩾ \sqrt{\frac{a^{2} + 1}{b^{2} + 1}} + \sqrt{\frac{b^{2} + 1}{c^{2} + 1}} + \sqrt{\frac{c^{2} + 1}{a^{2} + 1}}

for a, b, c are positive real number

Answered by EDWIN88 last updated on 09/Aug/21

(∙) w i t h C a u c h y - {S c h w a r z}^{'} s

(a^{2} + b^{2}) \sqrt{(a^{2} + 1) (b^{2} + 1)} ⩾ (a^{2} + b^{2}) (a b + 1)

= a b (a^{2} + b^{2}) + a^{2} + b^{2}

⩾ a b (a^{2} + b^{2} + 2)

\Rightarrow Σ \frac{a}{b} + Σ \frac{b}{a} = Σ \frac{a^{2} + b^{2}}{a b}

⩾ Σ \frac{a^{2} + b^{2} + 2}{\sqrt{(a^{2} + 1) (b^{2} + 1)}}

= Σ \sqrt{\frac{a^{2} + 1}{b^{2} + 1}} + Σ \sqrt{\frac{b^{2} + 1}{a^{2} + 1}}

(∙) w i t h {C h e b y s h e v}^{'} s

Σ \frac{a^{2}}{b^{2}} = Σ \frac{a^{2}}{b^{2} + 1} + Σ \frac{a^{2}}{b^{2} (b^{2} + 1)}

⩾ Σ \frac{a^{2}}{b^{2} + 1} + Σ \frac{b^{2}}{b^{2} (b^{2} + 1)}

= Σ \frac{a^{2} + 1}{b^{2} + 1}

t h e n {(1 + Σ \frac{a}{b})}^{2} = 1 + 2 (Σ \frac{a}{b} + Σ \frac{b}{a}) + \frac{a^{2}}{b^{2}}

⩾ 1 + 2 (Σ \sqrt{\frac{a^{2} + 1}{b^{2} + 1}} + Σ \sqrt{\frac{b^{2} + 1}{a^{2} + 1}}) + Σ \frac{a^{2} + 1}{b^{2} + 1}

= {(1 + Σ \sqrt{\frac{a^{2} + 1}{b^{2} + 1}})}^{2}

T h u s \frac{a}{b} + \frac{b}{c} + \frac{c}{a} ⩾ \sqrt{\frac{a^{2} + 1}{b^{2} + 1}} + \sqrt{\frac{b^{2} + 1}{c^{2} + 1}} + \sqrt{\frac{c^{2} + 1}{a^{2} + 1}}

Facebook Tweet Pin