# show the inequality (a raised to(c-a))*(b raised to(a-b))*(c raised to (b-c))=or<1.a,,b,c>0 a=or<b=or<c

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If use logarithms, you should be able to do the first step. Use of logarithms helps you to reverse the exponentiation.

log(a^(c-a)*b^(a-b)*c^(b-c))=<log 1

Use property of logarithms and transform the product in a sum of logarithms.

log(a^(c-a))+ log(b^(a-b))+ log(c^(b-c))=<0 (use log1=0)

Use one of the four logarithmic properties: the log of the power is equal to the log of the base multiplied by power.

(c-a)log a + (a-b)log b+ (b-c) log c=<0

Use a<b<c => log a< log b< log c( because logarithmic function is monotonically increasing)

Use c-a>=a-b and add a both sides => c>=2a-b.

Use 2a=<b+c

Use Cebishev's inequality to (c-a)log a + (a-b)log b=< (c-b)(log a)/2+ (c-b)(log b)/2 = (c-b)(log ab)/2

Use the logarithmic property: the log of the power is equal to the log of the base multiplied by power:(c-b)(log ab)/2 = (c-b)(log sqrt(ab))

(c-a)log a + (a-b)log b+ (b-c) log c=<(c-b)(log sqrt(ab))+ (b-c) log c=< 0

The right side hand is (c-b)(log sqrt(ab))+ (b-c) log c=(c-b)(log sqrt(ab))- (c-b) log c= (c-b)log (sqrt(ab)/c) =< 0

Check each factor of product (c-b)log (sqrt(ab)/c) is is positive or negative: b<c => 0<c-b =>c-b positive

sqrt(ab)<c=>log (sqrt(ab)/c) => (c-b)log (sqrt(ab)/c) =< 0

Answer: Inequality (c-b)log (sqrt(ab)/c) =< 0 proves (a^(c-a)*b^(a-b)*c^(b-c))=<1.