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


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

$A b o u t t h e s o l u t i o n t o q u e s t i o n :$ $F o r a, b, c > 0 a n d a b c = 1, p r o v e$ $a^{b + c} b^{c + a} c^{a + b} ⩽ 1$ $W a y 1 :$ ${L e t}^{'} s s a y a ⩽ b ⩽ c .$ $W e c a n p r o v e t h a t a ⩽ 1 :$ $I f a > 1, w e w i l l g e t b ⩾ a > 1, c ⩾ b > 1,$ $\Rightarrow a b c > 1$ $b u t a b c = 1!$ $s o a > 1 i s n o t t r u e, i . e . a ⩽ 1 .$ $S i m i l a r l y w e c a n a l s o p r o v e t h a t c ⩾ 1 :$ $I f c < 1, w e w i l l g e t b ⩽ c < 1, a ⩽ b < 1,$ $\Rightarrow a b c < 1$ $b u t a b c = 1$ $s o c < 1 i s n o t t r u e, i . e . c ⩾ 1 .$ $W e k n o w a l s o$ $i f p ⩽ 1, t h e n p^{x} ⩽ 1 f o r x ⩾ 0$ $i f p ⩾ 1, t h e n p^{x} ⩾ 1 f o r x ⩾ 0$ $S = a^{b + c} b^{c + a} c^{a + b} = a^{b + c} {(\frac{1}{a c})}^{c + a} c^{a + b}$ $= \frac{a^{b - a}}{c^{c - b}}$ $s i n c e a ⩽ 1 a n d b - a ⩾ 0, w e h a v e$ $a^{b - a} ⩽ 1$ $s i n c e c ⩾ 1 a n d c - b ⩾ 0, w e h a v e$ $c^{b - a} ⩾ 1$ $\Rightarrow S = \frac{a^{b - a}}{c^{c - b}} = \frac{⩽ 1}{⩾ 1} ⩽ 1$ $W a y 2 :$ $S = a^{b + c} b^{c + a} c^{a + b} = \frac{a^{a + b + c} b^{c + a + b} c^{a + b + c}}{a^{a} b^{b} c^{c}}$ $= \frac{{(a b c)}^{a + b + c}}{a^{a} b^{b} c^{c}} = \frac{1}{a^{a} b^{b} c^{c}} = \frac{1}{a^{a} b^{b} {(\frac{1}{a b})}^{\frac{1}{a b}}}$ $= \frac{{(a b)}^{\frac{1}{a b}}}{a^{a} b^{b}}$ ${l e t}^{'} s l o o k a t f u n c t i o n F (x, y) = \frac{{(x y)}^{\frac{1}{x y}}}{x^{x} y^{y}},$ $t h e g r a p h o f F (x, y) s e e c o m m e n t .$ $I t h a s a m a x i m u m a t (1, 1) w h i c h$ $i s F_{m a x} = 1 .$ $H e n c e f o r x, y > 0, 0 < F (x, y) ⩽ 1$ $\Rightarrow S = \frac{{(a b)}^{\frac{1}{a b}}}{a^{a} b^{b}} = F (a, b) = F (b, a) ⩽ 1$
Commented bymrW1 last updated on 18/May/17

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

$t h a n k s f o r e f f o r t s$
Commented bymrW1 last updated on 19/May/17

$F (x, y) = \frac{{(x y)}^{\frac{1}{x y}}}{x^{x} y^{y}} i s r e a l l y a n i n t e r e s t i n g$ $f u n c t i o n y o u c a n p l a y w i t h . . .$
Commented bymrW1 last updated on 19/May/17

Commented bymrW1 last updated on 19/May/17

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