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Question-146817




Question Number 146817 by mathdanisur last updated on 15/Jul/21
Answered by mindispower last updated on 15/Jul/21
⇔((y(x+y)+z(x+z))/( (√((x+y)(x+z)))))≥y+z....  y(x+y)+z(x+z)=x(y+z)+y^2 +z^2 ≥x(y+z)+(1/2)(y+z)^2   =(y+z)(x+(1/2)(y+z))=(y+z)((1/2)((x+y)+(x+z)))  ≥(y+z).(1/2).2(√((x+y)(x+z)))=(y+z)(√((x+y)(x+z)))  ⇒((y(x+y)+z(x+z))/( (√((x+y)(x+z)))))≥(((y+z)(√((x+y)(x+z))))/( (√((x+y)(x+z)))))=y+z
y(x+y)+z(x+z)(x+y)(x+z)y+z.y(x+y)+z(x+z)=x(y+z)+y2+z2x(y+z)+12(y+z)2=(y+z)(x+12(y+z))=(y+z)(12((x+y)+(x+z)))(y+z).12.2(x+y)(x+z)=(y+z)(x+y)(x+z)y(x+y)+z(x+z)(x+y)(x+z)(y+z)(x+y)(x+z)(x+y)(x+z)=y+z
Commented by mathdanisur last updated on 16/Jul/21
thank you Ser
thankyouSer

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