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Question Number 176555 by Shrinava last updated on 21/Sep/22

ABCD−convex quadrilateral M∈Int(ABCD) , F−area , s−semiperimetr a , b , c−sides. Prove that: ((MA^4 )/b) + ((MB^4 )/c^4 ) + ((MC^4 )/d) + ((MD^4 )/a) ≥ ((2F^2 )/s)

$$\mathrm{ABCD}−\mathrm{convex}\:\mathrm{quadrilateral} \\ $$$$\mathrm{M}\in\mathrm{Int}\left(\mathrm{ABCD}\right)\:,\:\mathrm{F}−\mathrm{area}\:,\:\mathrm{s}−\mathrm{semiperimetr} \\ $$$$\mathrm{a}\:,\:\mathrm{b}\:,\:\mathrm{c}−\mathrm{sides}.\:\mathrm{Prove}\:\mathrm{that}: \\ $$$$\frac{\mathrm{MA}^{\mathrm{4}} }{\mathrm{b}}\:+\:\frac{\mathrm{MB}^{\mathrm{4}} }{\mathrm{c}^{\mathrm{4}} }\:+\:\frac{\mathrm{MC}^{\mathrm{4}} }{\mathrm{d}}\:+\:\frac{\mathrm{MD}^{\mathrm{4}} }{\mathrm{a}}\:\geqslant\:\frac{\mathrm{2F}^{\mathrm{2}} }{\mathrm{s}} \\ $$

Commented by mr W last updated on 21/Sep/22

a=AB, b=BC etc.?

$${a}={AB},\:{b}={BC}\:{etc}.? \\ $$

Commented by Shrinava last updated on 21/Sep/22

yes dear professor

$$\mathrm{yes}\:\mathrm{dear}\:\mathrm{professor} \\ $$

Answered by mr W last updated on 24/Sep/22

Commented by Shrinava last updated on 27/Sep/22

thank you dear professor so that′s it?

$$\mathrm{thank}\:\mathrm{you}\:\mathrm{dear}\:\mathrm{professor} \\ $$$$\mathrm{so}\:\mathrm{that}'\mathrm{s}\:\mathrm{it}? \\ $$

Commented by mr W last updated on 04/Oct/22

s=((a+b+c+d)/2) F=((MA×MB×sin α)/2)+((MB×MC×sin β)/2)+((MC×MD×sin γ)/2)+((MD×MA×sin δ)/2) ≤(1/2)(MA×MB+MB×MC+MC×MD+MD×MA) ≤(1/2)(((MA^2 +MB^2 )/2)+((MB^2 +MC^2 )/2)+((MC^2 +MD^2 )/2)+((MD^2 +MA^2 )/2)) =((MA^2 +MB^2 +MC^2 +MD^2 )/2) ((2F^2 )/s)≤(((MA^2 +MB^2 +MC^2 +MD^2 )^2 )/(a+b+c+d)) ≤((MA^4 )/b)+((MB^4 )/c)+((MC^4 )/d)+((MD^4 )/a) ∗) proved! ✓ ∗) see Chaucy−Schwarz Inequality

$${s}=\frac{{a}+{b}+{c}+{d}}{\mathrm{2}} \\ $$$${F}=\frac{{MA}×{MB}×\mathrm{sin}\:\alpha}{\mathrm{2}}+\frac{{MB}×{MC}×\mathrm{sin}\:\beta}{\mathrm{2}}+\frac{{MC}×{MD}×\mathrm{sin}\:\gamma}{\mathrm{2}}+\frac{{MD}×{MA}×\mathrm{sin}\:\delta}{\mathrm{2}} \\ $$$$\:\:\:\leqslant\frac{\mathrm{1}}{\mathrm{2}}\left({MA}×{MB}+{MB}×{MC}+{MC}×{MD}+{MD}×{MA}\right) \\ $$$$\:\:\:\leqslant\frac{\mathrm{1}}{\mathrm{2}}\left(\frac{{MA}^{\mathrm{2}} +{MB}^{\mathrm{2}} }{\mathrm{2}}+\frac{{MB}^{\mathrm{2}} +{MC}^{\mathrm{2}} }{\mathrm{2}}+\frac{{MC}^{\mathrm{2}} +{MD}^{\mathrm{2}} }{\mathrm{2}}+\frac{{MD}^{\mathrm{2}} +{MA}^{\mathrm{2}} }{\mathrm{2}}\right) \\ $$$$\:\:\:=\frac{{MA}^{\mathrm{2}} +{MB}^{\mathrm{2}} +{MC}^{\mathrm{2}} +{MD}^{\mathrm{2}} }{\mathrm{2}} \\ $$$$\frac{\mathrm{2}{F}^{\mathrm{2}} }{{s}}\leqslant\frac{\left({MA}^{\mathrm{2}} +{MB}^{\mathrm{2}} +{MC}^{\mathrm{2}} +{MD}^{\mathrm{2}} \right)^{\mathrm{2}} }{{a}+{b}+{c}+{d}} \\ $$$$\left.\:\:\:\:\:\:\:\leqslant\frac{{MA}^{\mathrm{4}} }{{b}}+\frac{{MB}^{\mathrm{4}} }{{c}}+\frac{{MC}^{\mathrm{4}} }{{d}}+\frac{{MD}^{\mathrm{4}} }{{a}}\:\:\:\:\:\:\ast\right) \\ $$$$\:\:\:\:\:\:\:{proved}!\:\checkmark \\ $$$$ \\ $$$$\left.\ast\right)\:\:{see}\:{Chaucy}−{Schwarz}\:{Inequality} \\ $$

Commented by Shrinava last updated on 24/Sep/22

thank you dear professor please

$$\mathrm{thank}\:\mathrm{you}\:\mathrm{dear}\:\mathrm{professor}\:\mathrm{please} \\ $$

Commented by mr W last updated on 04/Oct/22

yes. now the proof is complete.

$${yes}.\:{now}\:{the}\:{proof}\:{is}\:{complete}. \\ $$

Commented by mr W last updated on 04/Oct/22

Commented by Shrinava last updated on 08/Oct/22

perfect dear professor thank you

$$\mathrm{perfect}\:\mathrm{dear}\:\mathrm{professor}\:\mathrm{thank}\:\mathrm{you} \\ $$

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