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Question Number 40344 by behi83417@gmail.com last updated on 20/Jul/18

Commented by MJS last updated on 20/Jul/18

maybe we can show p≥(3/2)

$$\mathrm{maybe}\:\mathrm{we}\:\mathrm{can}\:\mathrm{show}\:{p}\geqslant\frac{\mathrm{3}}{\mathrm{2}} \\ $$

Commented by behi83417@gmail.com last updated on 21/Jul/18

let:        x=tgA,y=tgB,z=tgC  now:   tgA+tgB+tgC=tgA.tgB.tgC⇒          A+B+C=180  it is just an idia.

$$\boldsymbol{{let}}:\:\:\:\:\:\:\:\:{x}={tgA},{y}={tgB},{z}={tgC} \\ $$$$\boldsymbol{{now}}:\:\:\:{tgA}+{tgB}+{tgC}={tgA}.{tgB}.{tgC}\Rightarrow \\ $$$$\:\:\:\:\:\:\:\:{A}+{B}+{C}=\mathrm{180} \\ $$$${it}\:{is}\:{just}\:{an}\:{idia}. \\ $$

Answered by MJS last updated on 20/Jul/18

it′s not true  x=1  y=4  z=(5/3)  p=((283571)/(51714))>4

$$\mathrm{it}'\mathrm{s}\:\mathrm{not}\:\mathrm{true} \\ $$$${x}=\mathrm{1} \\ $$$${y}=\mathrm{4} \\ $$$${z}=\frac{\mathrm{5}}{\mathrm{3}} \\ $$$${p}=\frac{\mathrm{283571}}{\mathrm{51714}}>\mathrm{4} \\ $$

Commented by behi83417@gmail.com last updated on 20/Jul/18

dear MJS!please consider the first  condition: x+y+z=xyz.

$${dear}\:{MJS}!{please}\:{consider}\:{the}\:{first} \\ $$$${condition}:\:\boldsymbol{{x}}+\boldsymbol{{y}}+\boldsymbol{{z}}=\boldsymbol{{xyz}}. \\ $$

Commented by MJS last updated on 20/Jul/18

I did:  1+4+(5/3)=((20)/3)  1×4×(5/3)=((20)/3)

$$\mathrm{I}\:\mathrm{did}: \\ $$$$\mathrm{1}+\mathrm{4}+\frac{\mathrm{5}}{\mathrm{3}}=\frac{\mathrm{20}}{\mathrm{3}} \\ $$$$\mathrm{1}×\mathrm{4}×\frac{\mathrm{5}}{\mathrm{3}}=\frac{\mathrm{20}}{\mathrm{3}} \\ $$

Commented by MrW3 last updated on 20/Jul/18

x+y+z=xyz  ⇒z=((x+y)/(xy−1))  p=...=F(x,y)  F(x,y) has no maxima.  p=F(1,4) =5.48...>4  p=F(1,100)=4900.5...>4  p=F(1,1000)=499000.5...>4    statement that p<q (=4 or what ever)  is not true.

$${x}+{y}+{z}={xyz} \\ $$$$\Rightarrow{z}=\frac{{x}+{y}}{{xy}−\mathrm{1}} \\ $$$${p}=...={F}\left({x},{y}\right) \\ $$$${F}\left({x},{y}\right)\:{has}\:{no}\:{maxima}. \\ $$$${p}={F}\left(\mathrm{1},\mathrm{4}\right)\:=\mathrm{5}.\mathrm{48}...>\mathrm{4} \\ $$$${p}={F}\left(\mathrm{1},\mathrm{100}\right)=\mathrm{4900}.\mathrm{5}...>\mathrm{4} \\ $$$${p}={F}\left(\mathrm{1},\mathrm{1000}\right)=\mathrm{499000}.\mathrm{5}...>\mathrm{4} \\ $$$$ \\ $$$${statement}\:{that}\:{p}<{q}\:\left(=\mathrm{4}\:{or}\:{what}\:{ever}\right) \\ $$$${is}\:{not}\:{true}. \\ $$

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