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Question Number 193116 by Mastermind last updated on 04/Jun/23
Show that for all a,b,c,d ∈ R with a,b,c,d ≥ 0 1) (√(ab))(√(cd)) ≤ (1/4)(a^2 +b^2 +c^2 +d^2 ) 2) (abcd)^(1/4) ≤ (1/4)(a+b+c+d) Help!
$$\mathrm{Show}\:\mathrm{that}\:\mathrm{for}\:\mathrm{all}\:\mathrm{a},\mathrm{b},\mathrm{c},\mathrm{d}\:\in\:\mathbb{R}\:\mathrm{with} \\ $$$$\mathrm{a},\mathrm{b},\mathrm{c},\mathrm{d}\:\geqslant\:\mathrm{0}\: \\ $$$$\left.\mathrm{1}\right)\:\sqrt{\mathrm{ab}}\sqrt{\mathrm{cd}}\:\leqslant\:\frac{\mathrm{1}}{\mathrm{4}}\left(\mathrm{a}^{\mathrm{2}} +\mathrm{b}^{\mathrm{2}} +\mathrm{c}^{\mathrm{2}} +\mathrm{d}^{\mathrm{2}} \right) \\ $$$$\left.\mathrm{2}\right)\:\left(\mathrm{abcd}\right)^{\frac{\mathrm{1}}{\mathrm{4}}} \:\leqslant\:\frac{\mathrm{1}}{\mathrm{4}}\left(\mathrm{a}+\mathrm{b}+\mathrm{c}+\mathrm{d}\right) \\ $$$$ \\ $$$$\mathrm{Help}! \\ $$
Answered by Subhi last updated on 04/Jun/23
1) 4(a^2 +b^2 +c^2 +d^2 )≥(a+b+c+d)^2 a+b+c+d≥4^4 (√(abcd)) (AM - GM) (a+b+c+d)^2 ≥16(√(abcd)) 4(a^2 +b^2 +c^2 +d^2 )≥16(√(abcd)) (1/4)(a^2 +b^2 +c^2 +d^2 )≥(√(ab)).(√(cd))
$$\left.\mathrm{1}\right)\:\mathrm{4}\left({a}^{\mathrm{2}} +{b}^{\mathrm{2}} +{c}^{\mathrm{2}} +{d}^{\mathrm{2}} \right)\geqslant\left({a}+{b}+{c}+{d}\right)^{\mathrm{2}} \\ $$$${a}+{b}+{c}+{d}\geqslant\mathrm{4}^{\mathrm{4}} \sqrt{{abcd}}\:\:\:\:\:\:\left({AM}\:-\:{GM}\right) \\ $$$$\left({a}+{b}+{c}+{d}\right)^{\mathrm{2}} \geqslant\mathrm{16}\sqrt{{abcd}} \\ $$$$\mathrm{4}\left({a}^{\mathrm{2}} +{b}^{\mathrm{2}} +{c}^{\mathrm{2}} +{d}^{\mathrm{2}} \right)\geqslant\mathrm{16}\sqrt{{abcd}} \\ $$$$\frac{\mathrm{1}}{\mathrm{4}}\left({a}^{\mathrm{2}} +{b}^{\mathrm{2}} +{c}^{\mathrm{2}} +{d}^{\mathrm{2}} \right)\geqslant\sqrt{{ab}}.\sqrt{{cd}} \\ $$
Commented by Subhi last updated on 04/Jun/23
4(a^2 +b^2 +c^2 +d^2 )≥(a+b+c+d)^2 (a+b+c+d)^2 =a^2 +b^2 +c^2 +d^2 +2cd+2ab+2ac+2ad+2bc+2bd = a^2 +b^2 +c^2 +d^2 +2(ab+ac+ad+bc+bd+cd) (i) a^2 +b^2 +c^2 +d^2 a^2 +b^2 ≥2(√(a^2 .b^2 ))=2ab (AM−GM) b^2 +c^2 ≥2bc ⇛ c^2 +d^2 ≥2cd a^2 +d^2 ≥2ad ⇛ b^2 +d^2 ≥2bd a^2 +c^2 ≥2ac sum the 5 equations 3(a^2 +b^2 +c^2 +d^2 )≥2(ab+bc+ac+ad+bd+cd) (ii) from (i) , (ii) (a+b+c+d)^2 ≤4(a^2 +b^2 +c^2 +d^2 )
$$\mathrm{4}\left({a}^{\mathrm{2}} +{b}^{\mathrm{2}} +{c}^{\mathrm{2}} +{d}^{\mathrm{2}} \right)\geqslant\left({a}+{b}+{c}+{d}\right)^{\mathrm{2}} \\ $$$$\left({a}+{b}+{c}+{d}\right)^{\mathrm{2}} ={a}^{\mathrm{2}} +{b}^{\mathrm{2}} +{c}^{\mathrm{2}} +{d}^{\mathrm{2}} +\mathrm{2}{cd}+\mathrm{2}{ab}+\mathrm{2}{ac}+\mathrm{2}{ad}+\mathrm{2}{bc}+\mathrm{2}{bd} \\ $$$$=\:{a}^{\mathrm{2}} +{b}^{\mathrm{2}} +{c}^{\mathrm{2}} +{d}^{\mathrm{2}} +\mathrm{2}\left({ab}+{ac}+{ad}+{bc}+{bd}+{cd}\right)\:\:\:\left({i}\right) \\ $$$${a}^{\mathrm{2}} +{b}^{\mathrm{2}} +{c}^{\mathrm{2}} +{d}^{\mathrm{2}} \\ $$$${a}^{\mathrm{2}} +{b}^{\mathrm{2}} \geqslant\mathrm{2}\sqrt{{a}^{\mathrm{2}} .{b}^{\mathrm{2}} }=\mathrm{2}{ab}\:\:\:\:\:\left({AM}−{GM}\right) \\ $$$${b}^{\mathrm{2}} +{c}^{\mathrm{2}} \geqslant\mathrm{2}{bc}\:\:\:\:\:\:\:\:\:\:\Rrightarrow\:\:{c}^{\mathrm{2}} +{d}^{\mathrm{2}} \geqslant\mathrm{2}{cd} \\ $$$${a}^{\mathrm{2}} +{d}^{\mathrm{2}} \geqslant\mathrm{2}{ad}\:\:\:\:\:\:\:\:\Rrightarrow\:\:{b}^{\mathrm{2}} +{d}^{\mathrm{2}} \geqslant\mathrm{2}{bd} \\ $$$${a}^{\mathrm{2}} +{c}^{\mathrm{2}} \geqslant\mathrm{2}{ac} \\ $$$${sum}\:{the}\:\mathrm{5}\:{equations} \\ $$$$\mathrm{3}\left({a}^{\mathrm{2}} +{b}^{\mathrm{2}} +{c}^{\mathrm{2}} +{d}^{\mathrm{2}} \right)\geqslant\mathrm{2}\left({ab}+{bc}+{ac}+{ad}+{bd}+{cd}\right)\:\:\:\left({ii}\right) \\ $$$${from}\:\left({i}\right)\:,\:\left({ii}\right) \\ $$$$\left({a}+{b}+{c}+{d}\right)^{\mathrm{2}} \leqslant\mathrm{4}\left({a}^{\mathrm{2}} +{b}^{\mathrm{2}} +{c}^{\mathrm{2}} +{d}^{\mathrm{2}} \right) \\ $$
Commented by Mastermind last updated on 04/Jun/23
Thank you so much
$$\mathrm{Thank}\:\mathrm{you}\:\mathrm{so}\:\mathrm{much} \\ $$
Commented by Mastermind last updated on 04/Jun/23
Thank you my boss
$$\mathrm{Thank}\:\mathrm{you}\:\mathrm{my}\:\mathrm{boss} \\ $$
Commented by Mastermind last updated on 04/Jun/23
What′s AM−GM?
$$\mathrm{What}'\mathrm{s}\:\mathrm{AM}−\mathrm{GM}? \\ $$
Commented by Subhi last updated on 04/Jun/23
Arithmetic geometric mean inequality its proof is below ↓
$${Arithmetic}\:{geometric}\:{mean}\:{inequality} \\ $$$${its}\:{proof}\:{is}\:{below}\:\downarrow \\ $$
Answered by Subhi last updated on 04/Jun/23
proof for AM − GM ((√a)−(√b))^2 ≥0 a+b−2(√(ab)) ≥0 ((a+b)/2)≥(√(ab)) ((a_1 +a_2 +.........a_n )/n)≥^n (√(a_1 .a_2 ......a_n )) ((a+b+c+d)/4)≥^4 (√(abcd))
$${proof}\:{for}\:{AM}\:−\:{GM} \\ $$$$\left(\sqrt{{a}}−\sqrt{{b}}\right)^{\mathrm{2}} \geqslant\mathrm{0} \\ $$$${a}+{b}−\mathrm{2}\sqrt{{ab}}\:\geqslant\mathrm{0} \\ $$$$\frac{{a}+{b}}{\mathrm{2}}\geqslant\sqrt{{ab}} \\ $$$$\frac{{a}_{\mathrm{1}} +{a}_{\mathrm{2}} +………{a}_{{n}} }{{n}}\geqslant^{{n}} \sqrt{{a}_{\mathrm{1}} .{a}_{\mathrm{2}} ……{a}_{{n}} } \\ $$$$\frac{{a}+{b}+{c}+{d}}{\mathrm{4}}\geqslant^{\mathrm{4}} \sqrt{{abcd}} \\ $$

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