Determine-minimum-value-of-sec-4-tan-2-sec-4-tan-2-over-all-kpi-2-and-k-Z-

Question Number 119372 by bobhans last updated on 24/Oct/20

D e t e r m i n e m i n i m u m v a l u e o f

\frac{\sec^{4} α}{\tan^{2} β} + \frac{\sec^{4} β}{\tan^{2} α}, o v e r a l l α, β \neq \frac{k π}{2}

a n d k \in Z

Answered by 1549442205PVT last updated on 24/Oct/20

S = \frac{\sec^{4} α}{\tan^{2} β} + \frac{\sec^{4} β}{\tan^{2} α} = \frac{1}{\cos^{4} α} . \frac{\cos^{2} β}{\sin^{2} β} + \frac{1}{\cos^{4} β} . \frac{\cos^{2} α}{\sin^{2} α}

= \frac{{(1 + \tan^{2} α)}^{2}}{\tan^{2} β} + \frac{{(1 + \tan^{2} β)}^{2}}{\tan^{2} α}

Applying Cauchy - Shwardz we have

S ⩾ \frac{{(2 + \tan^{2} α + \tan^{2} β)}^{2}}{\tan^{2} α + \tan^{2} β}

= \frac{4 + {(\tan^{2} α + \tan^{2} β)}^{2} + 4 (\tan^{2} α + \tan^{2} β)}{\tan^{2} α + \tan^{2} β}

= 4 + [\tan^{2} α + \tan^{2} β + \frac{4}{\tan^{2} α + \tan^{2} β}]

= 4 + {(\sqrt{\tan^{2} α + \tan^{2} β} - \frac{2}{\sqrt{\tan^{2} α + \tan^{2} β}})}^{2}

+ 4 ⩾ 8 . The equality ocurrs if and only

if {\begin{cases} \sqrt{\tan^{2} α + \tan^{2} β} = \frac{2}{\sqrt{\tan^{2} α + \tan^{2} β}} (1) \\ \frac{1 + \tan^{2} α}{\tan^{2} β} = \frac{1 + \tan^{2} β}{\tan^{2} α} (2) \end{cases}

(2) \Rightarrow \tan^{4} α + \tan^{2} α = \tan^{4} β + \tan^{2} β

\Leftrightarrow (\tan^{2} α - \tan^{2} β) (\tan^{2} α + \tan^{2} β + 1) = 0 (*)

(1) \Rightarrow \tan^{2} α + \tan^{2} β = 2 . Replace into

(*) we get \tan^{2} α = \tan^{2} β = 1

Thus, S = \frac{\sec^{4} α}{\tan^{2} β} + \frac{\sec^{4} β}{\tan^{2} α} has smallest

value equal to 8 when \tan^{2} α = \tan^{2} β = 1

Answered by bemath last updated on 24/Oct/20

s e t \to {\begin{cases} a = \tan^{2} α \\ b = \tan^{2} β \end{cases}

i t s u f f i c e s t o d e t e r m i n e t h e m i n i m u m v a l u e o f \frac{{(a + 1)}^{2}}{b} + \frac{{(b + 1)}^{2}}{a}

w i t h a, b ⩾ 0 . w e h a v e \frac{{(a + 1)}^{2}}{b} + \frac{{(b + 1)}^{2}}{a} = \frac{a^{2} + 2 a + 1}{b} + \frac{b^{2} + 2 b + 1}{a}

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

⩾ 4 \sqrt[4]{\frac{a^{2}}{b} . \frac{1}{b} . \frac{b^{2}}{a} . \frac{1}{a}} + 4 \sqrt{\frac{a}{b} . \frac{b}{a}} = 8

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