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1-Let-a-b-and-c-real-number-such-that-ab-a-b-1-3-bc-b-c-1-4-and-ac-a-c-1-5-Find-the-value-of-24abc-ab-ac-bc-2-Let-p-and-q-be-two-real-number-that-satisfy-p-

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Question Number 116323 by bobhans last updated on 03/Oct/20
(1)Let a,b and c real number such that ((ab)/(a+b)) = (1/3), ((bc)/(b+c)) = (1/4) and ((ac)/(a+c)) = (1/5). Find the value of ((24abc)/(ab+ac+bc)) ? (2) Let p and q be two real number that satisfy p.q=2013. What is the minimum value of (p+q)^2 ?
$$\left(\mathrm{1}\right)\mathrm{Let}\:\mathrm{a},\mathrm{b}\:\mathrm{and}\:\mathrm{c}\:\mathrm{real}\:\mathrm{number}\:\mathrm{such}\:\mathrm{that}\: \\ $$$$\frac{\mathrm{ab}}{\mathrm{a}+\mathrm{b}}\:=\:\frac{\mathrm{1}}{\mathrm{3}},\:\frac{\mathrm{bc}}{\mathrm{b}+\mathrm{c}}\:=\:\frac{\mathrm{1}}{\mathrm{4}}\:\mathrm{and}\:\frac{\mathrm{ac}}{\mathrm{a}+\mathrm{c}}\:=\:\frac{\mathrm{1}}{\mathrm{5}}.\:\mathrm{Find} \\ $$$$\mathrm{the}\:\mathrm{value}\:\mathrm{of}\:\frac{\mathrm{24abc}}{\mathrm{ab}+\mathrm{ac}+\mathrm{bc}}\:? \\ $$$$\left(\mathrm{2}\right)\:\mathrm{Let}\:\mathrm{p}\:\mathrm{and}\:\mathrm{q}\:\mathrm{be}\:\mathrm{two}\:\mathrm{real}\:\mathrm{number}\:\mathrm{that} \\ $$$$\mathrm{satisfy}\:\mathrm{p}.\mathrm{q}=\mathrm{2013}.\:\mathrm{What}\:\mathrm{is}\:\mathrm{the}\:\mathrm{minimum} \\ $$$$\mathrm{value}\:\mathrm{of}\:\left(\mathrm{p}+\mathrm{q}\right)^{\mathrm{2}} \:? \\ $$
Answered by john santu last updated on 03/Oct/20
(1) → { (((1/a)+(1/b)=3)),(((1/b)+(1/c)=4 )),(((1/a)+(1/c)=5)) :} summing the three equation ⇒2((1/a)+(1/b)+(1/c)) = 12 ⇒ (1/a)+(1/b)+(1/c) = 6 ; ((ab+ac+bc)/(abc)) = 6 ⇒ ((abc)/(ab+ac+bc)) = (1/6) , multiply both sides by 24 we get ((24abc)/(ab+ac+bc)) = 4
$$\left(\mathrm{1}\right)\:\rightarrow\begin{cases}{\frac{\mathrm{1}}{{a}}+\frac{\mathrm{1}}{{b}}=\mathrm{3}}\\{\frac{\mathrm{1}}{{b}}+\frac{\mathrm{1}}{{c}}=\mathrm{4}\:}\\{\frac{\mathrm{1}}{{a}}+\frac{\mathrm{1}}{{c}}=\mathrm{5}}\end{cases} \\ $$$$\:\:\:\:\:{summing}\:{the}\:{three}\:{equation} \\ $$$$\Rightarrow\mathrm{2}\left(\frac{\mathrm{1}}{{a}}+\frac{\mathrm{1}}{{b}}+\frac{\mathrm{1}}{{c}}\right)\:=\:\mathrm{12}\: \\ $$$$\Rightarrow\:\frac{\mathrm{1}}{{a}}+\frac{\mathrm{1}}{{b}}+\frac{\mathrm{1}}{{c}}\:=\:\mathrm{6}\:;\:\frac{{ab}+{ac}+{bc}}{{abc}}\:=\:\mathrm{6} \\ $$$$\Rightarrow\:\frac{{abc}}{{ab}+{ac}+{bc}}\:=\:\frac{\mathrm{1}}{\mathrm{6}}\:,\:{multiply}\:{both} \\ $$$${sides}\:{by}\:\mathrm{24}\:{we}\:{get}\:\frac{\mathrm{24}{abc}}{{ab}+{ac}+{bc}}\:=\:\mathrm{4} \\ $$$$ \\ $$
Answered by john santu last updated on 03/Oct/20
(2)(p+q)^2 =(p−q)^2 +4pq ≥ 0 + 4×2013 ≥ 8052 , when p=q=(√(2013)) Minimum value of (p+q)^2 equal to 8052
$$\left(\mathrm{2}\right)\left({p}+{q}\right)^{\mathrm{2}} =\left({p}−{q}\right)^{\mathrm{2}} +\mathrm{4}{pq}\:\geqslant\:\mathrm{0}\:+\:\mathrm{4}×\mathrm{2013} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\geqslant\:\mathrm{8052}\:,\:{when}\:{p}={q}=\sqrt{\mathrm{2013}} \\ $$$${Minimum}\:{value}\:{of}\:\left({p}+{q}\right)^{\mathrm{2}} \:{equal}\:{to}\:\mathrm{8052} \\ $$

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