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

Answered by ajfour last updated on 09/Oct/21

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

Commented by MathSh last updated on 10/Oct/21

$$\mathrm{Thank}\mathrm{you}\mathbf{S}\mathrm{er}$$

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