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Question Number 13333 by mrW1 last updated on 19/May/17

About the solution to question: For a,b,c>0 and abc=1, prove a^(b+c) b^(c+a) c^(a+b) ≤1 Way 1: Let′s say a≤b≤c. We can prove that a≤1: If a>1, we will get b≥a>1, c≥b>1, ⇒abc>1 but abc=1! so a>1 is not true, i.e. a≤1. Similarly we can also prove that c≥1: If c<1, we will get b≤c<1, a≤b<1, ⇒abc<1 but abc=1 so c<1 is not true, i.e. c≥1. We know also if p≤1, then p^x ≤1 for x≥0 if p≥1, then p^x ≥1 for x≥0 S=a^(b+c) b^(c+a) c^(a+b) =a^(b+c) ((1/(ac)))^(c+a) c^(a+b) =(a^(b−a) /c^(c−b) ) since a≤1 and b−a≥0, we have a^(b−a) ≤1 since c≥1 and c−b≥0, we have c^(b−a) ≥1 ⇒S= (a^(b−a) /c^(c−b) )=((≤1)/(≥1))≤1 Way 2: S=a^(b+c) b^(c+a) c^(a+b) =((a^(a+b+c) b^(c+a+b) c^(a+b+c) )/(a^a b^b c^c )) =(((abc)^(a+b+c) )/(a^a b^b c^c ))=(1/(a^a b^b c^c ))=(1/(a^a b^b ((1/(ab)))^(1/(ab)) )) =(((ab)^(1/(ab)) )/(a^a b^b )) let′s look at function F(x,y)=(((xy)^(1/(xy)) )/(x^x y^y )), the graph of F(x,y) see comment. It has a maximum at (1,1) which is F_(max) =1. Hence for x, y>0, 0<F(x,y)≤1 ⇒S=(((ab)^(1/(ab)) )/(a^a b^b ))=F(a,b)=F(b,a)≤1

$${About}\:{the}\:{solution}\:{to}\:{question}: \\ $$ $${For}\:{a},{b},{c}>\mathrm{0}\:{and}\:{abc}=\mathrm{1},\:{prove} \\ $$ $${a}^{{b}+{c}} {b}^{{c}+{a}} {c}^{{a}+{b}} \leqslant\mathrm{1} \\ $$ $$ \\ $$ $${Way}\:\mathrm{1}: \\ $$ $${Let}'{s}\:{say}\:{a}\leqslant{b}\leqslant{c}. \\ $$ $${We}\:{can}\:{prove}\:{that}\:{a}\leqslant\mathrm{1}: \\ $$ $${If}\:{a}>\mathrm{1},\:{we}\:{will}\:{get}\:{b}\geqslant{a}>\mathrm{1},\:{c}\geqslant{b}>\mathrm{1}, \\ $$ $$\Rightarrow{abc}>\mathrm{1} \\ $$ $${but}\:{abc}=\mathrm{1}!\: \\ $$ $${so}\:{a}>\mathrm{1}\:{is}\:{not}\:{true},\:{i}.{e}.\:{a}\leqslant\mathrm{1}. \\ $$ $${Similarly}\:{we}\:{can}\:{also}\:{prove}\:{that}\:{c}\geqslant\mathrm{1}: \\ $$ $${If}\:{c}<\mathrm{1},\:{we}\:{will}\:{get}\:{b}\leqslant{c}<\mathrm{1},\:{a}\leqslant{b}<\mathrm{1}, \\ $$ $$\Rightarrow{abc}<\mathrm{1} \\ $$ $${but}\:{abc}=\mathrm{1} \\ $$ $${so}\:{c}<\mathrm{1}\:{is}\:{not}\:{true},\:{i}.{e}.\:{c}\geqslant\mathrm{1}. \\ $$ $$ \\ $$ $${We}\:{know}\:{also}\: \\ $$ $${if}\:{p}\leqslant\mathrm{1},\:{then}\:{p}^{{x}} \leqslant\mathrm{1}\:{for}\:{x}\geqslant\mathrm{0} \\ $$ $${if}\:{p}\geqslant\mathrm{1},\:{then}\:{p}^{{x}} \geqslant\mathrm{1}\:{for}\:{x}\geqslant\mathrm{0} \\ $$ $$ \\ $$ $${S}={a}^{{b}+{c}} {b}^{{c}+{a}} {c}^{{a}+{b}} ={a}^{{b}+{c}} \left(\frac{\mathrm{1}}{{ac}}\right)^{{c}+{a}} {c}^{{a}+{b}} \\ $$ $$=\frac{{a}^{{b}−{a}} }{{c}^{{c}−{b}} } \\ $$ $${since}\:{a}\leqslant\mathrm{1}\:{and}\:{b}−{a}\geqslant\mathrm{0},\:{we}\:{have} \\ $$ $${a}^{{b}−{a}} \leqslant\mathrm{1} \\ $$ $${since}\:{c}\geqslant\mathrm{1}\:{and}\:{c}−{b}\geqslant\mathrm{0},\:{we}\:{have} \\ $$ $${c}^{{b}−{a}} \geqslant\mathrm{1} \\ $$ $$\Rightarrow{S}=\:\frac{{a}^{{b}−{a}} }{{c}^{{c}−{b}} }=\frac{\leqslant\mathrm{1}}{\geqslant\mathrm{1}}\leqslant\mathrm{1} \\ $$ $$ \\ $$ $${Way}\:\mathrm{2}: \\ $$ $${S}={a}^{{b}+{c}} {b}^{{c}+{a}} {c}^{{a}+{b}} =\frac{{a}^{{a}+{b}+{c}} {b}^{{c}+{a}+{b}} {c}^{{a}+{b}+{c}} }{{a}^{{a}} {b}^{{b}} {c}^{{c}} } \\ $$ $$=\frac{\left({abc}\right)^{{a}+{b}+{c}} }{{a}^{{a}} {b}^{{b}} {c}^{{c}} }=\frac{\mathrm{1}}{{a}^{{a}} {b}^{{b}} {c}^{{c}} }=\frac{\mathrm{1}}{{a}^{{a}} {b}^{{b}} \left(\frac{\mathrm{1}}{{ab}}\right)^{\frac{\mathrm{1}}{{ab}}} } \\ $$ $$=\frac{\left({ab}\right)^{\frac{\mathrm{1}}{{ab}}} }{{a}^{{a}} {b}^{{b}} } \\ $$ $${let}'{s}\:{look}\:{at}\:{function}\:{F}\left({x},{y}\right)=\frac{\left({xy}\right)^{\frac{\mathrm{1}}{{xy}}} }{{x}^{{x}} {y}^{{y}} }, \\ $$ $${the}\:{graph}\:{of}\:{F}\left({x},{y}\right)\:{see}\:{comment}. \\ $$ $$ \\ $$ $${It}\:{has}\:{a}\:{maximum}\:{at}\:\left(\mathrm{1},\mathrm{1}\right)\:{which} \\ $$ $${is}\:{F}_{{max}} =\mathrm{1}. \\ $$ $${Hence}\:{for}\:{x},\:{y}>\mathrm{0},\:\mathrm{0}<{F}\left({x},{y}\right)\leqslant\mathrm{1} \\ $$ $$\Rightarrow{S}=\frac{\left({ab}\right)^{\frac{\mathrm{1}}{{ab}}} }{{a}^{{a}} {b}^{{b}} }={F}\left({a},{b}\right)={F}\left({b},{a}\right)\leqslant\mathrm{1} \\ $$

Commented bymrW1 last updated on 18/May/17

Commented byAbbas-Nahi last updated on 19/May/17

thanks for efforts

$$\boldsymbol{{thanks}}\:\boldsymbol{{for}}\:\boldsymbol{{efforts}} \\ $$

Commented bymrW1 last updated on 19/May/17

F(x,y)=(((xy)^(1/(xy)) )/(x^x y^y )) is really an interesting function you can play with...

$${F}\left({x},{y}\right)=\frac{\left({xy}\right)^{\frac{\mathrm{1}}{{xy}}} }{{x}^{{x}} {y}^{{y}} }\:{is}\:{really}\:{an}\:{interesting} \\ $$ $${function}\:{you}\:{can}\:{play}\:{with}... \\ $$

Commented bymrW1 last updated on 19/May/17

Commented bymrW1 last updated on 19/May/17

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