Let a,b>0 and x∈]0;(π/2)[ Prove ((a/(sinx))+1)((b/(cosx))+1)≥(1+(√(2ab)))^2


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Question Number 117832 by snipers237 last updated on 13/Oct/20

$L e t a, b > 0 a n d x \in] 0; \frac{π}{2} [$ $P r o v e (\frac{a}{s i n x} + 1) (\frac{b}{c o s x} + 1) ⩾ {(1 + \sqrt{2 a b})}^{2}$
	Answered by 1549442205PVT last updated on 14/Oct/20
	$From the hypothesis a,b>0 and x∈]0;(π/2)[ we have sinx>0,cosx>0.Hence L.H.S=(a/(sinx))+(b/(cosx))+((ab)/(sinxcosx))+1 =((√(a/(sinx)))−(√(b/(cosx))))^2 +2(√((ab)/(sinxcosx))) +((2ab)/(sin2x))+1≥2(√((2ab)/(sin2x)))+((2ab)/(sin2x))+1 ≥2ab+2(√(2ab))+1=(1+(√(2ab)))^2 (due to 0<sin2x≤1).Hence,the inequality ((a/(sinx))+1)((b/(cosx))+1)≥(1+(√(2ab)))^2 is proved.The equality ocurrs if and only if { (( (a/(sinx))=(b/(cosx)))),((sin2x=1)) :} ⇔ { ((x=(π/4))),((a=b)) :}$
	$From the hypothesis a, b > 0 a n d x \in] 0; \frac{π}{2} [$ $we have sinx > 0, cosx > 0 . Hence$ $L . H . S = \frac{a}{sinx} + \frac{b}{cosx} + \frac{ab}{sinxcosx} + 1$ $= {(\sqrt{\frac{a}{sinx}} - \sqrt{\frac{b}{cosx}})}^{2} + 2 \sqrt{\frac{ab}{sinxcosx}}$ $+ \frac{2 ab}{\sin 2 x} + 1 ⩾ 2 \sqrt{\frac{2 ab}{\sin 2 x}} + \frac{2 ab}{\sin 2 x} + 1$ $⩾ 2 ab + 2 \sqrt{2 ab} + 1 = {(1 + \sqrt{2 ab})}^{2}$ $(due to 0 < \sin 2 x ⩽ 1) . Hence, the inequality$ $(\frac{a}{s i n x} + 1) (\frac{b}{c o s x} + 1) ⩾ {(1 + \sqrt{2 a b})}^{2}$ $is proved . The equality ocurrs if and$ $only if$ ${\begin{cases} \frac{a}{sinx} = \frac{b}{cosx} \\ \sin 2 x = 1 \end{cases} \Leftrightarrow {\begin{cases} x = \frac{π}{4} \\ a = b \end{cases}$
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