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

$$Aboutthesolutiontoquestion:$$$$Fora,b,c>0andabc=1,prove$$$$a^{b+c}{b}^{c+a}{c}^{a+b}⩽1$$$$$$$$Way1:$$$${Let}^{\prime }ssaya⩽b⩽c.$$$$Wecanprovethata⩽1:$$$$Ifa>1,wewillgetb⩾a>1,c⩾b>1,$$$$\Rightarrow abc>1$$$$butabc=1!$$$$soa>1isnottrue,i.e.a⩽1.$$$$Similarlywecanalsoprovethatc⩾1:$$$$Ifc<1,wewillgetb⩽c<1,a⩽b<1,$$$$\Rightarrow abc<1$$$$butabc=1$$$$soc<1isnottrue,i.e.c⩾1.$$$$$$$$Weknowalso$$$$ifp⩽1,then{p}^x⩽1forx⩾0$$$$ifp⩾1,then{p}^x⩾1forx⩾0$$$$$$$$S=a^{b+c}{b}^{c+a}{c}^{a+b}=a^{b+c}{\left(\frac{1}{ac}\right)}^{c+a}{c}^{a+b}$$$$=\frac{a^{b-a}}{c^{c-b}}$$$$sincea⩽1andb-a⩾0,wehave$$$$a^{b-a}⩽1$$$$sincec⩾1andc-b⩾0,wehave$$$$c^{b-a}⩾1$$$$\Rightarrow S=\frac{a^{b-a}}{c^{c-b}}=\frac{⩽1}{⩾1}⩽1$$$$$$$$Way2:$$$$S=a^{b+c}{b}^{c+a}{c}^{a+b}=\frac{a^{a+b+c}{b}^{c+a+b}{c}^{a+b+c}}{a^a{b}^b{c}^c}$$$$=\frac{{\left(abc\right)}^{a+b+c}}{a^a{b}^b{c}^c}=\frac{1}{a^a{b}^b{c}^c}=\frac{1}{a^a{b}^b{\left(\frac{1}{ab}\right)}^{\frac{1}{ab}}}$$$$=\frac{{\left(ab\right)}^{\frac{1}{ab}}}{a^a{b}^b}$$$${let}^{\prime }slookatfunctionF(x,y)=\frac{{\left(xy\right)}^{\frac{1}{xy}}}{x^x{y}^y},$$$$thegraphofF(x,y)seecomment.$$$$$$$$Ithasamaximumat(1,1)which$$$$is{F}_{max}=1.$$$$Henceforx,y>0,0<F(x,y)⩽1$$$$\Rightarrow S=\frac{{\left(ab\right)}^{\frac{1}{ab}}}{a^a{b}^b}=F(a,b)=F(b,a)⩽1$$

Commented by mrW1 last updated on 18/May/17

Commented by Abbas-Nahi last updated on 19/May/17

$$\mathit{t}\mathit{h}\mathit{a}\mathit{n}\mathit{k}\mathit{s}\mathit{f}\mathit{o}\mathit{r}\mathit{e}\mathit{f}\mathit{f}\mathit{o}\mathit{r}\mathit{t}\mathit{s}$$

Commented by mrW1 last updated on 19/May/17

$$F(x,y)=\frac{{\left(xy\right)}^{\frac{1}{xy}}}{x^x{y}^y}isreallyaninteresting$$$$functionyoucanplaywith\dots$$

Commented by mrW1 last updated on 19/May/17

Commented by mrW1 last updated on 19/May/17

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