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If-x-y-0-pi-2-Then-prove-that-log-sinx-2-sin2x-sinx-cosx-log-cosx-2-sin2x-sinx-cosx-2-

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Question Number 220072 by hardmath last updated on 04/May/25
If x,y∈(0,(π/2)) Then prove that: log_(sinx) ^2 (((sin2x)/(sinx + cosx))) + log_(cosx) ^2 (((sin2x)/(sinx + cosx))) ≥ 2
$$\mathrm{If}\:\:\:\mathrm{x},\mathrm{y}\in\left(\mathrm{0},\frac{\pi}{\mathrm{2}}\right) \\ $$$$\mathrm{Then}\:\mathrm{prove}\:\mathrm{that}: \\ $$$$\mathrm{log}_{\boldsymbol{\mathrm{sinx}}} ^{\mathrm{2}} \:\left(\frac{\mathrm{sin2x}}{\mathrm{sinx}\:+\:\mathrm{cosx}}\right)\:+\:\mathrm{log}_{\boldsymbol{\mathrm{cosx}}} ^{\mathrm{2}} \:\left(\frac{\mathrm{sin2x}}{\mathrm{sinx}\:+\:\mathrm{cosx}}\right)\:\geqslant\:\mathrm{2} \\ $$
Answered by MrGaster last updated on 05/May/25
sin 2x=2 sin x cos x⇒((sin 2x)/(sin x+cos x))=((2 sin x cos x)/(sin x+cos x)) log_(sin x) (((2 sin x cos x)/(sin x+cos x)))=((ln(((2 sin x cos x)/(sin x+cos x))))/(ln sin x)) log_(cos x) (((2 sin x cos x)/(sin x+cos x)))=((ln(((2 sin+cos x)/(sin x+cos x))))/(ln cos x)) a=log_(sin z) (((2 sin x cos x)/(sin x+cos x))),b=log_(cos x) (((2 sin x cos x)/(sin x+cos x))) a^2 +b^2 =(((ln A)/(ln sin x)))^2 +(((ln A)/(ln cos x)))^2 ,A=((2 sin x cos x)/(sin x+cos x)) (a^2 +b^2 )((1/a^2 )+(1/b^2 ))≥(1+1)^2 =4 Let u=ln sin x,v=ln sin (u,v<0) ln A=ln(2 sin x cos x)−ln(sin x+cos x)=ln 2+u++v−ln(sin x+cos x) x=(π/4)⇒sin x=cos x=((√2)/2)⇒A=((2∙((√2)/2)∙((√2)/2))/( (√2))) log_((√2)/2) ((1/( (√2))))=1⇒a=b=1⇒a^2 +b^2 =2 x≠(π/4)⇒sin x≠cos x⇒a^2 +b^2 >2 To sum up the originall formua holds if and only if x=(π/4) takes the equal sign.
$$\mathrm{sin}\:\mathrm{2}{x}=\mathrm{2}\:\mathrm{sin}\:{x}\:\mathrm{cos}\:{x}\Rightarrow\frac{\mathrm{sin}\:\mathrm{2}{x}}{\mathrm{sin}\:{x}+\mathrm{cos}\:{x}}=\frac{\mathrm{2}\:\mathrm{sin}\:{x}\:\mathrm{cos}\:{x}}{\mathrm{sin}\:{x}+\mathrm{cos}\:{x}} \\ $$$$\mathrm{log}_{\mathrm{sin}\:{x}} \left(\frac{\mathrm{2}\:\mathrm{sin}\:{x}\:\mathrm{cos}\:{x}}{\mathrm{sin}\:{x}+\mathrm{cos}\:{x}}\right)=\frac{\mathrm{ln}\left(\frac{\mathrm{2}\:\mathrm{sin}\:{x}\:\mathrm{cos}\:{x}}{\mathrm{sin}\:{x}+\mathrm{cos}\:{x}}\right)}{\mathrm{ln}\:\mathrm{sin}\:{x}} \\ $$$$\mathrm{log}_{\mathrm{cos}\:{x}} \left(\frac{\mathrm{2}\:\mathrm{sin}\:{x}\:\mathrm{cos}\:{x}}{\mathrm{sin}\:{x}+\mathrm{cos}\:{x}}\right)=\frac{\mathrm{ln}\left(\frac{\mathrm{2}\:\mathrm{sin}+\mathrm{cos}\:{x}}{\mathrm{sin}\:{x}+\mathrm{cos}\:{x}}\right)}{\mathrm{ln}\:\mathrm{cos}\:{x}} \\ $$$${a}=\mathrm{log}_{\mathrm{sin}\:{z}} \left(\frac{\mathrm{2}\:\mathrm{sin}\:{x}\:\mathrm{cos}\:{x}}{\mathrm{sin}\:{x}+\mathrm{cos}\:{x}}\right),{b}=\mathrm{log}_{\mathrm{cos}\:{x}} \left(\frac{\mathrm{2}\:\mathrm{sin}\:{x}\:\mathrm{cos}\:{x}}{\mathrm{sin}\:{x}+\mathrm{cos}\:{x}}\right) \\ $$$${a}^{\mathrm{2}} +{b}^{\mathrm{2}} =\left(\frac{\mathrm{ln}\:{A}}{\mathrm{ln}\:\mathrm{sin}\:{x}}\overset{\mathrm{2}} {\right)}+\left(\frac{\mathrm{ln}\:{A}}{\mathrm{ln}\:\mathrm{cos}\:{x}}\right)^{\mathrm{2}} ,{A}=\frac{\mathrm{2}\:\mathrm{sin}\:{x}\:\mathrm{cos}\:{x}}{\mathrm{sin}\:{x}+\mathrm{cos}\:{x}} \\ $$$$\left({a}^{\mathrm{2}} +{b}^{\mathrm{2}} \right)\left(\frac{\mathrm{1}}{{a}^{\mathrm{2}} }+\frac{\mathrm{1}}{{b}^{\mathrm{2}} }\right)\geq\left(\mathrm{1}+\mathrm{1}\right)^{\mathrm{2}} =\mathrm{4} \\ $$$$\mathrm{Let}\:{u}=\mathrm{ln}\:\mathrm{sin}\:{x},{v}=\mathrm{ln}\:\mathrm{sin}\:\left({u},{v}<\mathrm{0}\right) \\ $$$$\mathrm{ln}\:{A}=\mathrm{ln}\left(\mathrm{2}\:\mathrm{sin}\:{x}\:\mathrm{cos}\:{x}\right)−\mathrm{ln}\left(\mathrm{sin}\:{x}+\mathrm{cos}\:{x}\right)=\mathrm{ln}\:\mathrm{2}+{u}++{v}−\mathrm{ln}\left(\mathrm{sin}\:{x}+\mathrm{cos}\:{x}\right) \\ $$$${x}=\frac{\pi}{\mathrm{4}}\Rightarrow\mathrm{sin}\:{x}=\mathrm{cos}\:{x}=\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}\Rightarrow{A}=\frac{\mathrm{2}\centerdot\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}\centerdot\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}}{\:\sqrt{\mathrm{2}}} \\ $$$$\mathrm{log}_{\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}} \left(\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}\right)=\mathrm{1}\Rightarrow{a}={b}=\mathrm{1}\Rightarrow{a}^{\mathrm{2}} +{b}^{\mathrm{2}} =\mathrm{2} \\ $$$${x}\neq\frac{\pi}{\mathrm{4}}\Rightarrow\mathrm{sin}\:{x}\neq\mathrm{cos}\:{x}\Rightarrow{a}^{\mathrm{2}} +{b}^{\mathrm{2}} >\mathrm{2} \\ $$$$\mathrm{To}\:\mathrm{sum}\:\mathrm{up}\:\mathrm{the}\:\mathrm{originall} \\ $$$$\mathrm{formua}\:\mathrm{holds}\:\mathrm{if}\:\mathrm{and}\:\mathrm{only}\:\mathrm{if}\:{x}=\frac{\pi}{\mathrm{4}} \\ $$$$\mathrm{takes}\:\mathrm{the}\:\mathrm{equal}\:\mathrm{sign}. \\ $$
Commented by hardmath last updated on 05/May/25
Amazing solution as always, thank you very much my valuable professor
$$ \\ $$Amazing solution as always, thank you very much my valuable professor

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