Menu Close

Show-that-for-all-a-b-R-i-ab-1-2-a-2-b-2-ii-a-b-2-2-a-2-b-2-iii-ab-1-2-a-b-for-a-b-0-such-that-a-and-b-have-square-roots-Help-

Tinku Tara 0 Comments
Pin




Question Number 193138 by Mastermind last updated on 04/Jun/23
Show that for all a,b∈R i) ab≤(1/2)(a^2 +b^2 ) ii) (((a+b)/2))^2 ≤(a^2 +b^2 ) iii) (√(ab))≤(1/2)(a+b), for a,b≥0 such that a and b have square roots. Help
$$\mathrm{Show}\:\mathrm{that}\:\mathrm{for}\:\mathrm{all}\:\mathrm{a},\mathrm{b}\in\mathbb{R} \\ $$$$\left.\mathrm{i}\right)\:\mathrm{ab}\leqslant\frac{\mathrm{1}}{\mathrm{2}}\left(\mathrm{a}^{\mathrm{2}} +\mathrm{b}^{\mathrm{2}} \right) \\ $$$$\left.\mathrm{ii}\right)\:\left(\frac{\mathrm{a}+\mathrm{b}}{\mathrm{2}}\right)^{\mathrm{2}} \leqslant\left(\mathrm{a}^{\mathrm{2}} +\mathrm{b}^{\mathrm{2}} \right) \\ $$$$\left.\mathrm{iii}\right)\:\sqrt{\mathrm{ab}}\leqslant\frac{\mathrm{1}}{\mathrm{2}}\left(\mathrm{a}+\mathrm{b}\right),\:\mathrm{for}\:\mathrm{a},\mathrm{b}\geqslant\mathrm{0}\:\mathrm{such}\:\mathrm{that} \\ $$$$\mathrm{a}\:\mathrm{and}\:\mathrm{b}\:\mathrm{have}\:\mathrm{square}\:\mathrm{roots}. \\ $$$$ \\ $$$$\mathrm{Help} \\ $$
Answered by Subhi last updated on 04/Jun/23
(i) (a−b)^2 ≥0 a^2 +b^2 −2ab≥0 ab≤(1/2)(a^2 +b^2 ) (ii) (a+b)^2 =a^2 +b^2 +2ab note that 2ab≤a^2 +b^2 a^2 +b^2 +2ab≤2(a^2 +b^2 ) (((a+b)^2 )/2)≤a^2 +b^2 (iii) ((√a)−(√b))^2 ≥0 a+b−2(√(ab)) ≥0 (√(ab))≤(1/2)(a+b)
$$\left({i}\right)\:\left({a}−{b}\right)^{\mathrm{2}} \geqslant\mathrm{0} \\ $$$${a}^{\mathrm{2}} +{b}^{\mathrm{2}} −\mathrm{2}{ab}\geqslant\mathrm{0} \\ $$$${ab}\leqslant\frac{\mathrm{1}}{\mathrm{2}}\left({a}^{\mathrm{2}} +{b}^{\mathrm{2}} \right) \\ $$$$\left({ii}\right)\:\left({a}+{b}\right)^{\mathrm{2}} ={a}^{\mathrm{2}} +{b}^{\mathrm{2}} +\mathrm{2}{ab} \\ $$$${note}\:{that}\:\mathrm{2}{ab}\leqslant{a}^{\mathrm{2}} +{b}^{\mathrm{2}} \\ $$$${a}^{\mathrm{2}} +{b}^{\mathrm{2}} +\mathrm{2}{ab}\leqslant\mathrm{2}\left({a}^{\mathrm{2}} +{b}^{\mathrm{2}} \right) \\ $$$$\frac{\left({a}+{b}\right)^{\mathrm{2}} }{\mathrm{2}}\leqslant{a}^{\mathrm{2}} +{b}^{\mathrm{2}} \\ $$$$\left({iii}\right)\:\left(\sqrt{{a}}−\sqrt{{b}}\right)^{\mathrm{2}} \geqslant\mathrm{0} \\ $$$${a}+{b}−\mathrm{2}\sqrt{{ab}}\:\geqslant\mathrm{0} \\ $$$$\sqrt{{ab}}\leqslant\frac{\mathrm{1}}{\mathrm{2}}\left({a}+{b}\right) \\ $$
Answered by aba last updated on 04/Jun/23
i)(a−b)^2 ≥0 ⇒ ((a^2 +b^2 )/2)≥ab ii)ab=(1/2)((a+b)^2 −(a^2 −b^2 ))≤((a^2 +b^2 )/2) ⇒ (((a^2 +b^2 ))/2)≤a^2 +b^2 iii)((√a)−(√b))^2 ≥0 ⇒ ab≤((a+b)/2)
$$\left.\mathrm{i}\right)\left(\mathrm{a}−\mathrm{b}\right)^{\mathrm{2}} \geqslant\mathrm{0}\:\Rightarrow\:\frac{\mathrm{a}^{\mathrm{2}} +\mathrm{b}^{\mathrm{2}} }{\mathrm{2}}\geqslant\mathrm{ab} \\ $$$$\left.\mathrm{ii}\right)\mathrm{ab}=\frac{\mathrm{1}}{\mathrm{2}}\left(\left(\mathrm{a}+\mathrm{b}\right)^{\mathrm{2}} −\left(\mathrm{a}^{\mathrm{2}} −\mathrm{b}^{\mathrm{2}} \right)\right)\leqslant\frac{\mathrm{a}^{\mathrm{2}} +\mathrm{b}^{\mathrm{2}} }{\mathrm{2}}\:\Rightarrow\:\frac{\left(\mathrm{a}^{\mathrm{2}} +\mathrm{b}^{\mathrm{2}} \right)}{\mathrm{2}}\leqslant\mathrm{a}^{\mathrm{2}} +\mathrm{b}^{\mathrm{2}} \\ $$$$\left.\mathrm{iii}\right)\left(\sqrt{\mathrm{a}}−\sqrt{\mathrm{b}}\right)^{\mathrm{2}} \geqslant\mathrm{0}\:\Rightarrow\:\mathrm{ab}\leqslant\frac{\mathrm{a}+\mathrm{b}}{\mathrm{2}} \\ $$$$ \\ $$

Pin

Leave a Reply

Your email address will not be published. Required fields are marked *