Question Number 117832 by snipers237 last updated on 13/Oct/20
![Let a,b>0 and x∈]0;(π/2)[ Prove ((a/(sinx))+1)((b/(cosx))+1)≥(1+(√(2ab)))^2](Q117832.png)
$$\left.{Let}\:{a},{b}>\mathrm{0}\:\:{and}\:{x}\in\right]\mathrm{0};\frac{\pi}{\mathrm{2}}\left[\:\right. \\ $$ $$\:\:{Prove}\:\:\:\left(\frac{{a}}{{sinx}}+\mathrm{1}\right)\left(\frac{{b}}{{cosx}}+\mathrm{1}\right)\geqslant\left(\mathrm{1}+\sqrt{\mathrm{2}{ab}}\right)^{\mathrm{2}} \\ $$ $$ \\ $$
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)) :}](Q117864.png)
$$\left.\mathrm{From}\:\mathrm{the}\:\mathrm{hypothesis}\:{a},{b}>\mathrm{0}\:\:{and}\:{x}\in\right]\mathrm{0};\frac{\pi}{\mathrm{2}}\left[\:\right. \\ $$ $$\mathrm{we}\:\mathrm{have}\:\mathrm{sinx}>\mathrm{0},\mathrm{cosx}>\mathrm{0}.\mathrm{Hence} \\ $$ $$\mathrm{L}.\mathrm{H}.\mathrm{S}=\frac{\mathrm{a}}{\mathrm{sinx}}+\frac{\mathrm{b}}{\mathrm{cosx}}+\frac{\mathrm{ab}}{\mathrm{sinxcosx}}+\mathrm{1} \\ $$ $$=\left(\sqrt{\frac{\mathrm{a}}{\mathrm{sinx}}}−\sqrt{\frac{\mathrm{b}}{\mathrm{cosx}}}\right)^{\mathrm{2}} +\mathrm{2}\sqrt{\frac{\mathrm{ab}}{\mathrm{sinxcosx}}} \\ $$ $$+\frac{\mathrm{2ab}}{\mathrm{sin2x}}+\mathrm{1}\geqslant\mathrm{2}\sqrt{\frac{\mathrm{2ab}}{\mathrm{sin2x}}}+\frac{\mathrm{2ab}}{\mathrm{sin2x}}+\mathrm{1} \\ $$ $$\geqslant\mathrm{2ab}+\mathrm{2}\sqrt{\mathrm{2ab}}+\mathrm{1}=\left(\mathrm{1}+\sqrt{\mathrm{2ab}}\right)^{\mathrm{2}} \\ $$ $$\left(\mathrm{due}\:\mathrm{to}\:\:\mathrm{0}<\mathrm{sin2x}\leqslant\mathrm{1}\right).\mathrm{Hence},\mathrm{the}\:\mathrm{inequality} \\ $$ $$\:\:\left(\frac{{a}}{{sinx}}+\mathrm{1}\right)\left(\frac{{b}}{{cosx}}+\mathrm{1}\right)\geqslant\left(\mathrm{1}+\sqrt{\mathrm{2}{ab}}\right)^{\mathrm{2}} \\ $$ $$\mathrm{is}\:\mathrm{proved}.\mathrm{The}\:\mathrm{equality}\:\mathrm{ocurrs}\:\mathrm{if}\:\mathrm{and} \\ $$ $$\mathrm{only}\:\mathrm{if}\: \\ $$ $$\begin{cases}{\:\frac{\mathrm{a}}{\mathrm{sinx}}=\frac{\mathrm{b}}{\mathrm{cosx}}}\\{\mathrm{sin2x}=\mathrm{1}}\end{cases}\:\Leftrightarrow\begin{cases}{\mathrm{x}=\frac{\pi}{\mathrm{4}}}\\{\mathrm{a}=\mathrm{b}}\end{cases} \\ $$