Question and Answers Forum

All Questions      Topic List

Algebra Questions

Previous in All Question      Next in All Question      

Previous in Algebra      Next in Algebra      

Question Number 156253 by MathSh last updated on 09/Oct/21

Answered by ajfour last updated on 09/Oct/21

let b=pa, c=qa p,q≥1 a^3 pq=1 L.H.S.−R.H.S. =a^3 (1+p)(p+q)(q+1) −2a(1+p+q)−2 =a^3 ((1/a^3 )+p+q+1)(p+q) −2a(1+p+q)−2 =1+a^3 (p+q)^2 +a(a^2 −2)(p+q) −2a−2 say p+q=t f(t)=a^3 t^2 +a(a^2 −2)t−2(a+1)+1 D=(((2−a^2 )/(2a^2 )))^2 +((2(a+1)−1)/a^3 ) as D>0 a being +ve L.H.S.−R.H.S.=f(t)>0

$$\mathrm{let}\:\:\mathrm{b}=\mathrm{pa},\:\:\mathrm{c}=\mathrm{qa} \\ $$$$\mathrm{p},\mathrm{q}\geqslant\mathrm{1} \\ $$$$\mathrm{a}^{\mathrm{3}} \mathrm{pq}=\mathrm{1} \\ $$$$\mathrm{L}.\mathrm{H}.\mathrm{S}.−\mathrm{R}.\mathrm{H}.\mathrm{S}. \\ $$$$\:\:=\mathrm{a}^{\mathrm{3}} \left(\mathrm{1}+\mathrm{p}\right)\left(\mathrm{p}+\mathrm{q}\right)\left(\mathrm{q}+\mathrm{1}\right) \\ $$$$\:\:\:\:\:\:\:−\mathrm{2a}\left(\mathrm{1}+\mathrm{p}+\mathrm{q}\right)−\mathrm{2} \\ $$$$\:\:=\mathrm{a}^{\mathrm{3}} \left(\frac{\mathrm{1}}{\mathrm{a}^{\mathrm{3}} }+\mathrm{p}+\mathrm{q}+\mathrm{1}\right)\left(\mathrm{p}+\mathrm{q}\right) \\ $$$$\:\:\:\:\:\:−\mathrm{2a}\left(\mathrm{1}+\mathrm{p}+\mathrm{q}\right)−\mathrm{2} \\ $$$$=\mathrm{1}+\mathrm{a}^{\mathrm{3}} \left(\mathrm{p}+\mathrm{q}\right)^{\mathrm{2}} +\mathrm{a}\left(\mathrm{a}^{\mathrm{2}} −\mathrm{2}\right)\left(\mathrm{p}+\mathrm{q}\right) \\ $$$$\:\:\:\:\:\:−\mathrm{2a}−\mathrm{2} \\ $$$$\mathrm{say}\:\:\mathrm{p}+\mathrm{q}=\mathrm{t} \\ $$$$\mathrm{f}\left(\mathrm{t}\right)=\mathrm{a}^{\mathrm{3}} \mathrm{t}^{\mathrm{2}} +\mathrm{a}\left(\mathrm{a}^{\mathrm{2}} −\mathrm{2}\right)\mathrm{t}−\mathrm{2}\left(\mathrm{a}+\mathrm{1}\right)+\mathrm{1} \\ $$$$\mathrm{D}=\left(\frac{\mathrm{2}−\mathrm{a}^{\mathrm{2}} }{\mathrm{2a}^{\mathrm{2}} }\right)^{\mathrm{2}} +\frac{\mathrm{2}\left(\mathrm{a}+\mathrm{1}\right)−\mathrm{1}}{\mathrm{a}^{\mathrm{3}} } \\ $$$$\mathrm{as}\:\:\mathrm{D}>\mathrm{0}\:\:\:\:\mathrm{a}\:\mathrm{being}\:+\mathrm{ve} \\ $$$$\mathrm{L}.\mathrm{H}.\mathrm{S}.−\mathrm{R}.\mathrm{H}.\mathrm{S}.=\mathrm{f}\left(\mathrm{t}\right)>\mathrm{0} \\ $$

Commented by MathSh last updated on 10/Oct/21

Thank you Ser

$$\mathrm{Thank}\:\mathrm{you}\:\boldsymbol{\mathrm{S}}\mathrm{er} \\ $$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com